Math & Stat Part 2 Art & sci

7.3 METHOD OF RATIONALIZATION: = = lim  b+z − b−z × b+z + b− z 
 z b+z + 
If the function in the limit involves a square z→0  b − z 
root or a trigonometric function, it may be possible
to simplify the expression by multiplying and (b + z)−(b − z) × 1
dividing by its rationalizing factor. = lim  
z→0  z b+z + b − z 

SOLVED EXAMPLES = lim  2z × 1 
 z b+z + b − z 
z→0

lim  1 +x − 1  2
x 
Ex. 1. Evaluate : x→0  = lim  b − z  ...[As z→0, z ≠ 0]

z→0 b+z +

lim  1+ x −1 2
 x  =
Soluton : x→0
b+0+ b−0

= lim  1+ x −1× 1+ x +1 = 2
 x 1+ x +1 2b
x→0

 1+ x −1  = 1
lim  1+ x +1  b
( ) = 
x→0  x 

 x2 + x − 20 
Ex. 3. Evaluate lim  
( )  x→4 
x  x2 − 7 − 25 − x2 

= lim  x 1+ x +1   x2 + x − 20 
x→0   
Solution : lim 
 x→4  x2 − 7 − 25 − x2 

lim 1
=  1+ x +1 ...[As x→0, x ≠ 0]
x→0  x2 + x − 20 × x2 − 7 + x2 
x2 − 7 − 25 − x2 x2 − 7 + 25 − x2 
11 = lim  25 −

x→4

= =
1+0 +1 2

( )=  ( 4) 
 x − ( x + 5) x 2 −7 + 25 − x2 
 x 2− − 25 + x2 
 (b + 1 − (b − 1  lim 7
 z 
z)2 z)2  x→4

Ex. 2. Evaluate lim 

z→0 

( )=  ( 4) x2 
 x − ( x + 5) x2 − −7 + 25 − x2 
 2( 16) 
 (b + 1 1  lim
 
z)2 −(b − z)2  x→4

Solution : lim z
z→0 

( )=  ( 4) x2 
 x − ( x + 5) 4)( −7 + 25 − x2 
 2( x x+ 4) 
=  b+z − b− z  lim −
lim  z 
z→0   x→4

142

( )=  
 ( x + 5) x2 − 7 + 25 − x2   x2 + 9 − 2x2 + 9 
 3x2 + 4 − 2x2 + 4 
lim 2(x + 4)  5. = lim 

x→4 x→0

...[As x→4, x − 4 ≠ 0] Q.III Evaluate the Following limits :

( )(4 + 5)

=
42 − 7 + 25 − 42 = (9)(3 + 3) = 27 1. lim  x2 +x x − 2 
 x −1 
2(4 + 4) 2(8) 8 x→1  

EXERCISE 7.3 2.  1+ x2 − 1+ x 
lim  
x→0  1+ x3 − 1+ x 

Q.I Evaluate the following limits :  x2 + x − 20 

3. lim  3x + 4 − 4 

 6 + x + x2 − 6 x→4
lim  
1. x→0  x 

4. lim 3− 5+z 
 
z→4  1 − 5 − z 

2.  2x +3 − 4x −3 
lim  
x→3  x2 − 9  5. lim  3 − 1 
 9− x 
x→0 x x

3.  1− y2 − 1+ y2 
lim  
y→0  y2 
7.4 LIMIT OF A TRIGONOMETRIC
4.  2+x − 6− x  FUNCTION :
lim  x− 
x→2  Let’s Learn :
2 To evaluate the limits involving trigonometric
functions, we state -
Q.II Evaluate the following limits :
1) lim sinx = sina
1.  a + 2x − 3x  x→a
lim  
x→a  3a + x − 2 x  2) lim cosx = cosa
x→a
2. lim  x2 − 4 
  Using these results and trigonometric identities,
x→2  x+2− 3x − 2  we solve some examples.

3.  1+ 2+x − 3 Evaluation of limits can be done by the
lim   method of Factorization, Rationalization or
x→2  x−2  Simplification as the case may be. While solving
examples based on trigonometric functions we
 a+y− a can use trigonometric identities.
4. lim  
y→0  y a+y  Squeeze theorem (Also known as Sandwich
theorem)

143

Suppose f(x), g(x) and h(x) are given In ∆OAP,
functions such that f(x) ≤ g(x) ≤ h(x) for all x in PM
an open interval about a.
sin θ =
Suppose lim f(x) = L and lim h(x) = L OP

x→a x→a ∴ PM = OP sin θ
= r sin θ
So, lim f(x) ≤ lim g(x) ≤ lim h(x) Also, in ∆OAB

x→a x→a x→a AB
tan θ = OA
⇒ L≤ lim g(x) ≤ L ∴ lim g(x) = L ∴ AB = OA tan θ
= r tan θ
x→a x→a using these in (I), we get

7.4.2 Theorem : lim sinθ = 1; where θ is 1 11
θ r.r. sin θ < r2 θ < r.r tan θ
θ →0
2 22
measured in radian.

Proof : First consider the case when θ is
tending to zero through positive values.

We may take 0 < θ < π .
2

Draw a standard circle with radius r i.e. θ1 1
circles with centre at origin O and radius r.

Let A be the point of intersection of the i.e. 1 < sinθ < cosθ ... [Divide by r2 sin θ]
circle and the X-axis. Take point P on the circle 2
such that m∠AOP = θ
sinθ
Draw PM ⊥ OX. Draw a line through A ∴ 1 > θ > cos θ
parallel to Y-axis to meet OP extended at B
(fig. 7.2) i.e. cos θ < sinθ <1
θ

Area of ∆OAP < Area of sector OAP < Area Taking limit as θ → 0+
of ∆OAB

1 11 ∴ lim cos θ ≤ lim sinθ ≤ lim 1
∴ OA.PM< r2θ < OA.OB ..... (I) θ
θ →0+ θ →0+ θ →0+
2 22

∴ 1≤ lim sinθ ≤1
θ
θ →0+

By using squeeze theorem

∴ lim sinθ .... (II)
θ =1
θ →0+

Now suppose θ→0 through negative values

Let θ = – φ where φ > 0. Also as θ → 0, φ→0

∴ lim sinθ = lim sin(−φ )
θ
θ →0− φ →0+ −φ

= lim − sinφ = lim sin φ =1
φ→0+ −φ
Fig. 7.3 φ →0+ φ

144

∴ lim sinθ =1 .... (III) Ex. 2 : Evaluate : lim  cot x −1 
θ  
θ →0− x→π cosec2 x − 2
4

∴ from (II) and (III), lim sinθ =1 Solution : lim  cot x −1 
θ
θ →0

x→π  cosec2 x − 2 
4
Note:
x −1
Corollary 1 : lim  θ  = 1 = lim  cot +1− 2 
 sin   cot2 x 
θ →0 θ x→π
4

Corollary 2 : lim  tan θ  =1 = lim  cot x −1 
 θ   cot 2 x −1 
θ →0 x→π

4

Corollary 3 : lim  θ  =1 lim  cot x −1 
 tan θ   cot 
θ →0 = x→π ( + 1) ( 1)
4
cot x x −

Corollary 4 : lim  sin pθ  = 1 , (p constant.) 1
 pθ 
θ →0   = lim  

x→π ( cot x + 1)
4
 tan pθ  π
Corollary 5 : lim  pθ  =1, (p constant.) .....[As x 4 cot x − 1 ≠ 0]
 
θ →0

 pθ  11
 sin pθ  =
Corollary 6 : lim   =1, (p constant.) =  π  1+1
 4 
θ →0 cot +1

Corollary 7 : lim  pθ  =1, (p constant.) 1
 tan pθ  =
θ →0  
2

SOLVED EXAMPLES lim  3 − sin x − 2 
 cos2 x 
Ex. 3) Evaluate : x → 3π
2

Ex. 1) If 3x2 + 2 ≤ f(x) ≤ 5x2 − 6 for all x ∈ R, lim  3 − sin x − 2 
 cos2 x 
then find lim f(x). Solution : x → 3π
2
x→−2

Solution : Let g(x) = 3x2 + 2 and h(x) = 5x2 − 6 = lim  3 − sin x − 2 × 3 − sin x + 2 
 cos2 x 3 − sin x + 2 
So, we have g(x) ≤ f(x) ≤ h(x) x → 3π
2

Taking as limit x→ −2 throughout we = lim  3 − sin x − 4 
get  x 3 − sin x 
( ) x → 3π  cos2 +2 
2  

lim g(x) ≤ lim f(x) ≤ lim h(x)  −1− sin x 
x 3 − sin x 
x→−2 x→−2 x→−2  

lim (3x2+2) ≤ lim f(x) ≤ lim (5x2− 6) ( ) = lim  1 − sin 2 +2

x→−2 x→−2 x→−2 x → 3π
2

3(−2)2 + 2 ≤ lim f(x) ≤ 5(−2)2 − 6  − (1+ sin x) 
 x)(1+ sin x) 
x→−2  3 − sin 
 
( ) = lim
14 ≤ lim f(x) ≤ 14 (1− sin x + 2
x → 3π
x→−2 2

∴ lim f(x) = 14 ....... [By squeeze .....As x→, 3π , sin x → −1 and 1+sinx ≠ 0
2
x→−2

theorem]

145

=  −1  1× 8
  = 1× 4 = 2 (... as x → 0, 8x → 0, 4x → 0)
 
 1 − sin  3π    3 − sin  3π  +    sin pθ   tan pθ  =1
  2     2  2   pθ  pθ 
  .... lim   =1 lim  
 
θ →0 θ →0

= (1+1) −1 1 Ex. 6. Evaluate :  2sinx - sin2x 
3+1+ 2 =−  x3 
( ) lim
8
x→0

Solution : lim  2sinx − sin2 x 
 x3 
Ex. 4. Evaluate : lim  sin7θ  x→0
 θ 
θ →0

= lim  2sinx − 2sinx.cosx 
 x3 
Solution :  sin7θ  x→0
 θ 
lim  2sinx(1 − cosx) 
 x.x2 
θ →0

 sin 7θ  = lim
 7θ 
x→0

= lim =×17

θ →0 2sinx − cosx)
x x2
... as θ → 0, 7θ → 0 lim   × lim  (1 
   
= x→0 x→0

=1×7  lim sin pθ =   sinx   (1 − cosx) (1 + cosx) 
pθ 1  x   x2 (1 + cosx) 
θ →0 = 2 × lim × lim  × 

x→0 x→0

=7

 (1 − cos 2 x) 1 
 x2 cosx) 
 sin8x  = 2 × 1 × lim  × 
 tan 4x 
Ex. 5. Evaluate : lim x→0 (1 +

x→0

 sin8x  = 2 × lim  ( sin2 x)  × lim  1 
 tan4x   x2   cosx) 
Solution : lim x→0   x→0  (1 + 

x→0

Divide Numerator and Denominator by x  sinx  2  1 
 x   cos0) 
= 2 × lim × lim

 sin8x  x→0 x→0 (1 +

= lim  x x  = 2 (1)2 ×  1   lim sin θ = 1
 tan4  1 +  θ
x→0   θ →0

x 1

lim  sin8 x  1
 x  =2×1× =1
= x→0
2

lim  tan4 x 
 x 
x→0  sin x2 (1 − cos x2 ) 
 x6 
 sin8 x 8 Ex. 7. Evaluate : lim  
 8x
x→0

lim ×

= x→0  sin x2 (1− cos x2 ) 
x6 
lim  tan4 x × 4 Solution : lim  
 4x 
x→0 x→0

146

= lim  sin x2 (1− cos x2 ) × 1+ cos x2    πx     πx  
 x6 1+ cos x2    45     180  
x→0    sin  π  sin  π

=– 2 lim × × lim ×

= lim  sin x2 (1− cos2 x2) × 1  x→0  πx  45 x→0  πx  180
 x6 1+ cos x2   45   180 
x→0  
π π   sin pθ  
 sin x 2 . sin 2 x2 1  = −2 × (1) × 45 × (1) × 180 ......  lim  pθ  = 1
x6 ×  
= lim  θ →0 
1+ cos x2
x→0 

−2× 4× π × π π 2
  = 180 180 = −8  180
   
sin 3 x2 1  
× 
lim 3
( ) = 1+ cos x2
x→0  x2 Activity-1 lim tan x − sin x
 sin3 x
x→0

 x 2  3
 
= lim  sin 2  × lim  1 + 1 x 2  Solution : lim tan x − sin x
x  cos  x→0 sin3 x
x2 →0 x→0

.....[As x→ 0, x2 → 0]  1 
 x×� 
1 11 lim  sin sin3 − sin x 
==  x 
= (1)3 1+ cos 0 1+1 2 = x→0  

.....   sin θ  = 1, here θ = x 2   1 −� 
lθi→m0  θ    cos 
sin x

= lim x

x→0 sin3 x

lim  cos 5xο − cos 3xο 
Ex. 8. Evaluate :  x2  − cos x)
x→0   lim  (1 cos x × 1 
 sin 2 x 
= x→0

lim  cos 5xο − cos 3xο  (1 cos x) 1
 x2  cos x 1
Solution : x→0   = lim

x0

lim  −2sin 4xο sin xο  lim (1 cos x) 1
x2  co 1 cos )
= x→0   = x0


= – 2 lim  sin 4 xο × sin xο  = lim  1 × 1 
 x x   cos cos 
x→0   x→0 x (1 + x)

 sin 4  πx  sin  πx   = 1 1 
 180   180    × 
–  ×   cos� ) 
(1+ co� s
= 2 lim

x→0  x x 11
  ×
= � (1 +� )

 1× 1
 πx πx  = 1�
– 2 lim  sin  45  sin  180  
=      
x→0 × 1
= �
 x x 

147

EXERCISE 7.4 Activity-1:

Q.I Evaluate the following limits : lim 1− cos px = 1 p2
x2 2
x→0

1.  sin(mθ )  Consider,
 tan(nθ ) 
lim   lim 1− cos px

θ →0

xx→0 2

2. lim 1 − cos2θ 
 θ2 
θ →0 = lim 2.

x→0 x2

3. lim  x.tanx  Use 1– cos A = 2sin2 A
1− cosx  2
x→0

lim  sec x − 1  lim 2. 2
 x2 
4. x→0 = x→0 x

Q.II Evaluate the Following limits :  sin px 2
 2
1− cos(nx)  = lim 2 ×� 
1 − cos(mx)   px 
1. lim  x→0 

x→0 2 

2. lim  2 − cosecx   sin px 2  p2 
 cot2 x − 3   2
x→π lim 2 
6  px 
= x→0   
  4 

3. lim  cosx − sinx  2
 cos2x 
x→π
4

= 2 (1)2  p2  ∵ lim sin θ
  θ =1
Q.III Evaluate the following limits :  4  x→0

1.  cos (ax) − cos(bx)  =
lim  
x→0  cos (cx) −1 

2.  1− cosx − 2 lim 1− cos px = 1 p2
lim  sin2 x  x2 2
x→π   x→0

 tan 2 x − cot2 x  7.5 Substitution Method :
 secx − cosecx 
3. lim   We will consider examples of trigonometric

x→π functions in which x→a where generally a takes
4

4.  2sin2x + sinx −1  the values such as π, π , π , π , π etc. In such
 2sin2x − 3sinx +1 2 3 4 6
lim 
a case we put x – a = t so that as x→a, t→o.
x→π
6

148

SOLVED EXAMPLE = – 2 lim  sin  a + t  lim sin(t / 2)   1 
t→0   2   t / 2   2 
t→0

 1   sin θ  =1
 cos x  2  θ 
Ex. 1. Evaluate lim   = −2sin(a + 0).(1) .... lim
 π 
x→π θ →0
2
x −
 2
= −sina

Solution : Put x – π =t ∴x= π +t
2 2

π lim 1+ cosπ x 
2 Ex. 3. Evaluate  
As x→ ; t→0. x→1  (1− x)2 

   cos  π + t   Solution : Put 1 – x = t ∴ x = 1 – t;
 cos x    2   As x→1, t→0.
lim  = lim  
 π  t 
x→π  x −  t→0
2
2  
lim 1+ cosπ x = lim 1+ cos[π (1− t)]
   t 2 
x→1  (1− x)2  t→0

= lim  − sin t  1+ cos (π − π t ) 
 t 
t→0

sin t  = lim  (1− x)2 
 t   
t→0

= – lim 1− cosπ t 

t→0

= lim  t2 

lθi→m0 sin θ  t→0
θ 1
=–1   =   πt 
    2
 2 sin 2 

= lim  t2 

t→0 
 cos x − cos a 
Ex. 2. Evaluate lim    
 x−a 
x→a

 πt  2
 2
Solution : Put x – a = t ∴ x = a + t;  sin   
 
As x→a, t→0. = 2 lim

t→0 t 

lim  cos x − cos a  = lim  cos(a + t) − cos a   
 x−a   t 
x→a   t→0

  πt  2
  2
 sin t   π 2
  2 
 −2 sin  2a + t .sin t  2 lim π  
  2 2  = t→0 
 t 
= lim    2 

t→0

  π2    sin Pθ  =1
....  Pθ 
= 2(1)   lim

−  t . sin(t / 2)  4 θ →0
 2 t 
= 2 lim sin a +

t→0

= π2
2

149

lim  3 − tan x  4 (1)× 1 .......  lim tan θ = 1
Evaluate x→π 3 tan 0 θ
3 θ →0
Ex. 4. 3  π − 3x  ( )= 1+
 

Solution : Put π − x = t, ∴x= π – t, = 4
3 3 3

As x → π , t →0
3
EXERCISE 7.5


 3 − tan 
 3 − tan x    I) Evaluate the following
lim  π − 3x  = lim  π 
 3  3 − x  
x→π x→π  
3 3 

lim  cos ec x −1
 
 1) x→π   π − x 2 
2   2 

  π   
1   3   
= lim 3 − tan − t 

3 t→0 
 t  lim sin x − sin a
2) x→a 5 x − 5 a

 tan(π / 3) − tan t 
1 1+ tan(π / 3) tan t 
= lim  3 −   5 + cos x − 2 
3 t→0  t 
 3) lim  (π − x)2 
x→π  

lim  cos x − 3 sin x
   
  4) x→π  π −6x 
 3 − tan t  6
1 3 − + 3 tan t 
= lim
3 t→0  1

 t  5) lim  1− x2 
   
x→1  sin π x 

  II) Evaluate the following
 
1  3 + 3 tan t − 3 + tan t 

= lim 1+ 3 tan t
3 t→0  
 t  lim  2sin x −1
 
1) x→π  π −6x 
6

= 1 lim  4 tan t   2 − cos x − sin x 
3  + 3 tan  lim  
t→0  t (1 t )  2) x→π  (4x −π )2 
4

4 lim  tan t  lim  1 lim  2 − 3 cos x − sin x 
3  t    
= t→0 t→0  (1 + 3 tan t)  3) x→π  (6x −π )2 
 6

150

4) ª sin º SOLVED EXAMPLES
lim « x sin a »
xoa « x a
»
¬¼
ª5x 1º
Ex. 1. Evaluate : lim « »
cos 3x 3cos x ¬ sinx ¼
lim (2x )3 xo0

5) x ª5x 1º

2

Solution : lim « sinx »
¬ ¼
xo0

7.6 LIMITS OF EXPONENTIAL AND Divide Numerator and Denominator by x
LOGARITHMIC FUNCTIONS :
ª 5x 1º
« »
Let's :Learn = lim « x »
xo0 « sinx »

«¬ x ¼»

We use the following results without proof. ª5x 1º

§ e x 1 · lim « x »
¨ x ¸ ¬ ¼
1) lim © ¹ log e 1 = xo0

xo0 ª sinx º
«¬ x »¼
§ a x 1 · lim
¨ x ¸
© ¹ xo0

2) lim log a (a > 0, a ≠ 0)

xo0 log 5
= 1

1 ª § sin T · § a x 1 · º
« ¨© T ¸¹ ¨ x ¸ log a»
3) lim>1 x@x e ¬ © ¹
xo0 ¼

.... lim 1, lim

§ log 1 x · T o0 xo0

4) lim ¨ x ¸1 = (log5)
xo0 © ¹

5) lim § e px 1 · 1, (p constant)
¨ px ¸
xo0 © ¹ Ex. 2. Evaluate : lim ª 5x - 3x º
« x »
xo0 ¬ ¼

lim § a px 1 · log a , (p constant) ª 5x 3x º
¨ px ¸ x »
6) xo0 © ¹ Solution : Given lim « ¼
¬
xo0

7) § log 1 px · 1, (p constant) ª5x 1 3x 1º
lim ¨ ¸
xo0 © px ¹ = lim « x »
¬ ¼
xo0

1 ª 5x 1 (3x 1) º

8) lim 1 px px e , (p constant) = lim « »
xo0 «¬ »¼
xo0 x

= lim 5x 1 – lim 3x 1

xo0 x xo0 x

151

 ax −1 31
×
= log5 – log3 ..... lim   = log a  2  2 3
  lim 3x 
x→a x  x→0 1 + 3x  
 2  

= log  5  = −5×1
 3  23
 2 
lim 1 + −5x 
 x→0 2  −5x 
  

1

1 5x  x 3
6 
Ex. 3. Evaluate : lim + e6  (1 ) 1 e
kx
x→0 = −5 ..... lim + kx =

e6 x→0

1

lim 1 + 5x  x 84
6 
Solution : x→0 = e=6 e3

5 Ex. 5. Evaluate : lim  log 4 + log(0.25 + x) 
 x 
1 1 6 x→0

= lim + 5x  6 
6  5x 
x→0  log 4 + log(0.25 + x) 
 x 
Solution : lim

x→0

5  log [ 4(0.25 + x)]
 x
= (e)6  

= lim

x→0

1

 3x + 2  3 x  log (1 + 4x)
 2 −5x  
Ex. 4. Evaluate : lim = lim  x 

x→0 x→0

11  log (1 + 4 x ) 
 x 
lim  3x + 2  3x = lim  2 + 3x  3x = 4 × lim  4 
 2 −5x   2 − 5x 
Solution: x→0 x→0 x→0

1 = 4(1) .....  lim  log(1 + px)  
  px  = 1
 1 3x   3 x x→0
 2   
 2 + 

= lim 5x =4
2
x→0  2 1 −  
  
 e2 x + e−2x − 2 

1 Ex. 6. Evaluate : lim  x sin x 

x→0

 1 + 3x 1 3 Solution :
 2 
= lim  x  ( )lim
 5x  
x→0 2  x→0
 1 1   e2x + e−2x − 2  = lim  e2x e2x + e−2x − 2 
  x sin x   e2x x sin x 
− x   x→0  


= lim  e4x +1− 2e2x 
 e2x x sin x 
x→0  

152

= lim  (e2 x )2 − 2e2x + 1   (3x −1)(7x −1) 
 e2x x sin x 
x→0    x2 
= lim  x log(1+ x)  ....[ As x → 0, x2 ≠ 0]

x→0  
 
= lim  (e2x −1)2 × 1   x2
 x sin x e2x 
x→0    3x −1× 7 x− 
 x x 
 log(1 + x) 1 
 
 (e2x −1)2  = lim

 1 x→0
= lim 
x2 × e2x   x 
x→0  x sin x 
log 3.log 7
x2 =
1
....[As x→0, x ≠ 0, x2≠0]

 e2x  2  ax −1 log (1 + x) 
lim 2   x 1
 x→0  − 1   ×4 ....  lim = log a, lim x = 
 x 
  x→0 x→0

=  sin x  × lim  1 
 x   e2x 
lim x→0 = log 3.log 7

x→0

= (1) 2× 4 × 1 .....  lim ekx − 1 = 1, lim sin θ =  Activity-3
1 eο  kx θ 1
x→0 θ →0  8x − 4x − 2x +1

=4 Evaluate : lim  x2 
 
x→0

= lim  (4× � )x − 4x − 2x +1
x→0  
lim  21x − 7x − 3x +1  x2 
 
Ex. 7. Evaluate : x→0  x log(1+ x)  x 4x 2x 1
x2
 21x − 7x − 3x +1  4x
lim

x0

Solution : lim  x log(1+ x)  (2 1)
  x2
x→0 lim

x0

= lim  7x.3x − 7x − 3x +1  lim (2x 1). (4x 1)
  x2
x→0  x log(1+ x)  x0

= lim   7x (3x −1) − (3x −1)    (2x − 1)   (4x − 1) 
 x   x 
x→0  x log(1+ x)  = lim   × lim  

x→0 x→0

= lim  (3x −1)(7x −1)  =

x→0  x log(1+ x) 



153

Activity-4: Q.II Evaluate the following limits :

 e x − sin x −1  3x + 3−x − 2 

Evaluate : lim  x  1) lim  
  
x→0 x→0 x.tanx

 (ex −1) −�  1
 x 
= lim   2) lim  3 + x  x
 3 − x 
x→0 x→0

� sin x  2
 � 
= lim − lim  5x + 3  x
 3− 2x 
x→0 x 3) x→0

= lim  �  − lim  sin x   log (3 − x) − log(3 + x) 
   x 
x→0 x x→0 4) lim  
x→0  
x

= −1 1
= 1−1
5) lim 4x +1 x
1 − 4x 
x→0

= 1

 5 + 7x  3 x
 5 − 3x 
EXERCISE 7.6 6) lim

x→0

Q.I Evaluate the following limits : Q.III Evaluate the following limits :

1) lim 9x − 5x  1) lim  ax − bx 
   − sin 
x→0  4x −1  x→0  sin ( 4x) (2 x) 

5x + 3x − 2x −1  2x −1 3 
2) lim   ( )2) lim  
 x  x→0  (3x −1).sinx.log(1+ x) 
x→0



3) lim  ax + bx + cx − 3  15x − 5x − 3x +1
 sin x 
x→0   3) lim  
 
x→0 x . sinx

4) lim  6x + 5x + 4x − 3x+1   (25)x − 2(5)x +1
 
x→0  sin x  4) lim  

x→0  x . sinx 

5) lim  8sin x − 2tan x   (49)x − 2(35)x + (25)x 
 
x→0  e2x −1  5) lim  
x→0  sin x.log(1+ 2x) 

154

7.7 LIMIT AT INFINITY : Definition : A function f is said to tend to limit ‘l’
(FUNCTIONTENDINGTOINFINITY) as x tends to – ∞ if for given ∈ > 0, there exists
a positive number M such that |f(x) – l| < ∈, for
Let's :Learn all x > M

∴ lim f (x) = l
x→− ∞

7.7.1 Limit at infinity :

Let us consider the function f ( x) = 1 Note : Whenever expression is of the

x form ∞ , then divide, by suitable power of
∞

Observe that as x approaches to ∞ or –∞ the value x to get finite limits of numerator as well as

of f(x) is shown below, denominator.

i) Observe the following table for f (x) = 1
x
7.7.2 Infinite Limits :
x 1 10 100 1000 10000 100000 … Let us consider the function f(x) = 1 .
f(x) 1 0.1 0.01 0.001 0.0001 0.00001 … x

Observe the behavior of f(x) as x approaches zero

We see that as x assumes larger and larger from right and from left.
values, 1 assumes the value nearer and nearer
i) Observe the following table for f ( x) = 1
x
to zero. x

1 x= 1 0.1 0.01 0.001 0.0001 0.00001 …
∴ lim = 0 f(x) 1 10 100 1000 10000 100000 …

x→∞ x

Definition : A function f is said to tend to limit ‘l’ We see that as x assumes values nearer 0, but
as x tends to ∞ if for given ∈ > 0, there exists a greater than 0, 1 assumes the values larger and
positive number M such that |f(x) – l| < ∈, ∀ x in
the domain of f for which x > M x
larger.

1
lim f (x) = lim → ∞
∴ lim f (x) = l ∴ x→0+ xx→0+
x→∞

ii) Observe the following table for f (x) = 1 ii) Observe the following table for f ( x) = 1
x
x
x -1 -10 -100 -1000 -10000 -100000 …
f(x) -1 -0.1 -0.01 -0.001 -0.0001 -0.00001 … x = -1 -0.1 -0.01 -0.001 -0.0001 -0.00001 …
f(x) -1 -10 -100 -1000 -10000 -100000 …
We see that as x assumes values which tend
to −∞, 1 assumes the value nearer and nearer to We see that as x assumes values nearer to
0, but less than 0, 1 assumes the values which
x
zero. x
tends to – ∞

∴ 1 =0
lim
xx→−∞ lim f (x) = lim 1 → −∞
x→0−
∴ x→0− x

155

SOLVED EXAMPLES 10 + 0 + 0 1
= 5 − 0 + 0 ( lim = 0)

x→∞ x

Ex. 1. Evaluate : lim  ax +b  10
 cx + d  =
x→∞
5
Solution : lim  ax + b  = 2
 cx + d 
x→∞

 ax + b  Ex. 3. Evaluate : lim  x2 + 3x −� x 
x→∞  
 
= lim  cx x d  Solution : lim  x2 + 3x − x 
 +  x→∞  
x→∞

x

a b  ( x2 + 3x − x)( x2 + 3x + x) 
x  lim  
lim + = x→∞  ( x2 + 3x + x) 

= x→∞

lim c + d 
x 
x→∞  x2 + 3x − x2 
lim 
= x→∞  x2 + 3x + x 

a+0 1
= c+0 --- as x → ∞, x →0

 3x 
lim  
= a = x→∞ 
c x2 + 3x + x 

 3x 
 
lim  3 
Ex. 2. Evaluate : lim 10x2 +5x+3 = x→∞  x 1+ + x
   
x→∞  5x2 -3x+8  x

Solution : lim 10x2 + 5x + 3   
 − 3x + 8   
x→∞  5x2  3

Divide by x2 to get finite limits of the lim  3 
numerator as well as of the denominator, = x→∞  1+ +1
 x 

 10x2 + 5x + 3 
 
= lim  5x2 x2  3 1
x→∞  − 3x + 8  = 1+ 0 +1 ( lim = 0)

3 x→∞ x
= 1+1
 x2 
3
lim 10 + 5 + 3  = 2
x x2 
= x→∞

lim 5 − 3 + 8 
x x2 
x→∞

156

EXERCISE 7.7 Let's Remember

I Evaluate the following : Note : For limits of trigonometric functions,
angle is supposed to be in radian.
 ax3 + bx2 + cx + d 
1) lim  ex3 + fx2 + gx + h  1) lim sin x = 1
  x
x→∞ x→0

2) lim  ( x3 + 3x + 2  2) lim sin x = sin a
 (x 
x→∞  x + 4)(x − 6) − 3)  x→a

3) lim cos x = cos a

x→0

3) lim  7x2 + 5x − 3  4) lim tan x = 1
 8x2 − 2x + 7  x
x→∞   x→0

II Evaluate the following : 5) lim sin kx =k
x
x→0

1) lim  7 x2 + 2x −3  6) lim tan kx =k
 x4 + x + 2  x
x→∞   x→0

2) lim  x2 + 4x +16 − x2 + 16  7) lim x.sin 1 =0
x→∞    x 
x→0

3) lim  x4 + 4x2 − x 2  8) lim 1–cosp x = p2
x→∞   x2 2
x→0

III Evaluate the following : 9) lim cosm x– cosn x = n2 − m2
x2 2
x→0

1) lim  ( )( ) 3x2 + 4 4x2 − 6 (5x2 + 2)   10) lim ax–1 = log a, "a>0
x
4x6 + 2x4 −1 x→0

x→∞   lim 1

2)  ( 3x − 4)3 ( 4x + 3)4  11) x→0 (1+ x) x =e
 
lim  ( 3x + 2)7 

x→∞

12) lim log (1 + x) = 1
x
x→0

( )3)
lim  x x +1− x   xn − an 
x→∞  
lim
13)  x−a  = nan–1 for a > 0
x→a  
 ( )20 ( )30 
4) lim  2 x − 3x − 
x→∞  1 1 

( 2 x + )50  ex −1

1 lim

14) x→0  x  =1
 
 x2 + 5 − x2 − 3 
5) lim  
x→∞  x2 + 3 − x2 +1 
1
15) lim  x  =0

x→∞

157

16) lim k = 0 for k, p ∈ R and p > 0 lim 3cos x + cos 3x  =
 x p   
x→∞ 7) x→π  (2x −π )3 
2

1 A) 3 B) 1 C) – 1 D) 1
17) As x → 0,  x  → ∞ 2 2 2 4

 a x lim  15x − 3x − 5x +1 =
 b   sin2 x 
18) lim = 0, if a < b 8) x→0  

x→∞

A) log 15 B) log 3 + log 5

MISCELLANEOUS EXERCISE - 7 C) log 3.log 5 D) 3 log 5

1

 3 + 5 x  x =
 3 − 4 x 
I) Select the correct answer from the given 9) lim  
alternatives.
x→0

 x4 −16  A) e3 B) e6 C) e9 D) e–3
x2 −5x +6 
1) lim   =

x→2  log(5 + x) − log(5 − x) 
 sin x 
10) lim   =

A) 23 B) 32 C) –32 D) –16 x→0

2) lim  x7 +128  = A) 3 B) – 5 C) – 1 D) 2
x3 +8  2 2 2 5
x→−2  


56 112 121 28 
3 3 3 3  3cos x −1 
A) B) C) D) lim   =
 
11) x→π π −x
2
2 
 1 1  π
lim  −11x + −x  =
3)  x2 + 24 x2 −6  A) 1 B) log 3 C) 3 2 D) 3 log 3
x→3

A) − 2 B) 2 C) 7 D) − 7 lim  x.log (1+ 3x)  =
25 25 25 25 12)  
x→ 0  (e3x −1)2 

lim  x+4 −3  = A) 1 B) 1 C) 1 D) 1
 3x −11− 2  e9 e3 9 3
4) x→5

A) −2 B) 2 C) 5 D) 2 lim  (3sin x −1)3  =
9 7 9 9  
13) x→ 0  (3x −1) ⋅ tan x ⋅ log(1 + x) 

 tan 2 x − 3  A) 3 log 3 B) 2 log 3
 sec3 x − 8 
5) lim   = C) (log 3)2 D) (log 3)3

x→π
3

A) 1 B) 1 C) 1 D) 1 lim  5x−3 − 4x−3  =
2 3 4  
14) x→ 3  sin( x − 3) 

lim  5 sin x − x cos x  = A) log 5 – 4 B) log 5
2 tan x −3x2  4
6) x→0  
 log 5 log 5
C) log 4 D) 4
A) 0 B) 1 C) 2 D) 3

158

 (2x + 3)7 (x − 5)3  = lim  log x − log 2 
 x − 2 
15) lim  (2x − 5)10  12) x→2
 
x→ ∞

A) 3 B) 1 C) 1 D) 1 13) lim  abx − axb 
8 8 6 4  x2 −1 
x→1  

II) Evaluate the following.  5x −1 2 
( )14) lim  
(1− x)5 −1 x→0  (2x −1)log(1+ x) 
1) lim   
 (1 − x )3 −1
x→0

15) (2x +1)2 (7x − 3)3 
2) lim[x] ([*] is a greatest integer function.) lim  
x→0 x→∞  (5x + 2)5 

3) If f(r) = πr2 then find lim  f (r + h) − f (r)  x cos a − a cos x 
  x − a 
h→0  h  16) lim

x→a

4)  x x x2  lim  (sin x − cos x)2 
lim  +  17) x→π  
x→0   4  2 − sin x − cos x 

5) Find the limit of the function, if it exists, 18) lim  22x−2 − 2x +1
at x = 1 x→1  
 sin2 (x −1) 
7 − 4x for x < 1
f(x) = 
 x2 +2 for x ≥1  4x−1 − 2x +1
19) lim
 (x −1)2 
x→1  

6) Given that 7x ≤ f(x) ≤ 3x2 – 6 for all x.

Determine the value of lim f ( x) 20) lim  x −1
x→3  
x→1  log x 

7) lim  secx2 −1 
 x4 
x→0    1− cos x 
 x 
21) lim

x→0

8) lim  ex + e−x − 2 
 x.tanx 
x→0   22) lim x 3x2 5x3 (2n 1)xn n2
x→1 x1

9) lim  cos x(6x − 3x ) x) 
 − cos(4 
x→0  (6x)  23) lim

x→0

10) lim  a3x − a2x − ax +1  1   x2   x4   x2  x4   
 x.tanx  1  2  cos  4  cos  2  cos  4 
x→0    − cos − +    
  
x12   

11) lim  sinx − sina  24) lim  8x2 + 5x + 3  4 x+3
 x − a  x→∞ 8 x −1
x→a  
 
2x2 − 7x − 5

vvv

159

8 CONTINUITY

Let's Study 8.1.1 CONTINUITYOFAFUNCTIONATA
POINT
• Continuity of a function at a point.
• Continuity of a function over an interval. We are going to study continuity of functions of
• Intermediate value theorem. real variable so the domain will be an interval in
R. Before we consider a formal definition of a
Let's Recall function to be continuous at a point, let’s consider
various functions that fail to meet our notion of
• Different types of functions. continuity. The functions are indicated by graphs
• Limits of Algebraic, Trigonometric, where y = f(x)
Exponential and Logarithmic functions.
• Left hand and Right hand limits of functions.

8.1 CONTINUOUSANDDISCONTINUOUS Fig. 8.1 Fig. 8.2
FUNCTIONS
Fig. 8.3
The dictionary meaning of the word continuity is
the unbroken and consistent existence over a The function in figure 8.1 has a hole at x = a.
period of time. The intuitive idea of continuity is In fact f(x) is not defined at x = a.
manifested in the following examples.
The function in figure 8.2 has a break at x = a.
(i) An unbroken road between two cities.
For the function in figure 8.3, f(a) is not in the
(ii) Flow of river water. continuous line.

(iii) Railway tracks.

(iv) The changing temperature of a city during a
day.

In winter the temperature of Pune rises from
140C at night to 290C in the afternoon. This
change in the temp is continuous and all the
values between 14 and 29 are taken during 12
hours. An activity that takes place gradually,
without interruption or abrupt change is called a
continuous process. There are no jumps, breaks,
gaps or holes in the graph of the function.

160

8.1.2 DEFINITION OF CONTINUITY Hence f (x) is continuous at x = 0.

A function f(x) is said to be continuous at a point Illustration 2 : Consider f (x) = x2 and let us
x = a, if the following three conditions are discuss the continuity of f at x = 2.
satisfied:
f (x) = x2
i. f is defined at every point on an open interval
containing a. ∴ f (2) = 22 = 4

ii. lim f (x) exists lim f (x) = lim (x2) = 22 = 4
x→a
x→2 x→2
iii. lim f (x) = f (a).
x→a ∴ lim f (x) = f (2) = 4

Among the three graphs given above, decide x→2
which conditions of continuity are not satisfied.
∴ The function f (x) is continuous at x = 2.
The condition (iii) can be reformulated and the
continuity of f(x) at x = a, can be restated as Observe that f(x) = x3, x4, ... etc. are continuous
follows : at every point. It follows that all polynomials are
continuous functions of x.
A function f(x) is said to be continuous
at a point x = a if it is defined in some There are some functions, which are defined
neighborhood of ‘a’ and if in two different ways on either side of a point.
In such cases we have to consider the limits of
lim [f (a + h) −f (a)] = 0. function from left as well as right of that point.

h→0 8.1.3 CONTINUITY FROM THE RIGHT
AND FROM THE LEFT
Illustration 1. Let f (x) = |x| be defined on R.
A function f (x) is said to be continuous from the
f (x) = −x , for x < 0 right at x = a if lim f (x) = f (a).
= x , for x ≥ 0
xoa

A function f (x) is said to be continuous from the
left at x = a if lim f (x) = f (a).

xoa

If a function is continuous on the right and also
on the left of a then it is continuous at a because

lim f(x) = f(a) = lim f (x) .
xoa xoa

Fig. 8.4 Illustration 3: Consider the function f (x) = ¬«x»¼
in the interval [2, 4).
Consider, lim f (x) = lim (−x) = 0
xo0 x→0

lim f (x) = lim (x) = 0 Note : «¬x¼» is the greatest integer function or floor
xo0 x→0 function.

lim f (x) = lim f (x) = f (0) = 0
xo0 xo0

161

Solution : for x ∈ [ 2, 4) lim f (x) = lim (5x − 4.5) = 15 − 4.5 = 10.5
f (x) = ¬«x»¼ , for x ∈ [ 2, 3)
that is f (x) = 2 , for x ∈ [3, 4) xo3 x→3
= 3 ,
lim f (x) = lim f (x) = 10.5

xo3 xo3

∴ lim f (x) = f (3)

The graph of which is as shown in figure 8.5 x→3
Test of continuity at x = 3.
For x = 3, f(3) = 3 ∴ f(x) is continuous at x = 3.

Fig. 8.5

lim f (x) = lim «¬ x ¼» = lim (2) = 2, and

xo3 xo3 xo3

lim f (x) = lim «¬ x ¼» = lim (3) = 3 Fig. 8.6

xo3 xo3 xo3 8.1.4 Examples of Continuous Functions.

lim f (x) ≠ lim f (x) (1) Constant function, that is f(x) = k, is
xo3 xo3 continuous at every point on R.

∴ f (x) is discontinuous at x = 3. (2) Power functions, that is f(x) = xn, with
positive integral exponents are continuous at
Illustration 4 : every point on R.
3
(3) Polynomial functions,
Consider f (x) = x2 + for 0 ≤ x ≤ 3
2 P(x) = a0xn + a1xn-1+ a2xn-2 + ...... + an-1x + an
are continuous at every point on R
= 5x − 4.5 for 3 < x ≤ 5;
(4) The trigonometric functions sin x and cos x
3 are continuous at every point on R.
For x = 3, f (3) = 32 + = 10.5
(5) The exponential function ax (a > 0) and
2 logarithmic function logbx (for x, > 0, and b,

lim f (x) = lim (x2 + 3 3 = 10.5 b ≠ 1) are continuous on R.
) = 32 + 2
xo3 x→3
2

162

(6) Rational functions are of the form P(x) , Illustration 5: Consider
Q(x) f (x) = x2 − x − 5, for −4 ≤ x < − 2.
= x3 − 4x − 3, for −2 ≤ x ≤ 1.
Q(x) ≠ 0. They are continuous at every point For x = -2, f (−2) = (−2)3 − 4 (−2) −3 = −3
a if Q(a) ≠ 0.

8.1.5 PROPERTIES OF CONTINUOUS lim f (x) = lim (x2 − x − 5) = 4 + 2 − 5 = 1 and
FUNCTIONS: xo 2 xo 2

If the functions ƒ and g are continuous at x = a, lim f (x) = lim (x3 − 4x − 3) = −8 + 8 − 3 = −3
then, xo 2 xo 2

1. their sum, that is ( ƒ + g ) is continuous at ∴ lim f (x) ≠ lim f (x)
x = a. xo 2 xo 2

2. their difference, that is ( ƒ – g) or (g – f) is Hence lim f(x) does not exist.
continuous at x = a. xo 2

3. the constant multiple of f(x), that is k.ƒ, for ∴ the function f(x) has a jump discontinuity.
any k ∈ R, is continuous at x = a.

4. their product, that is ( ƒ.g ) is continuous at 8.1.8 REMOVABLE DISCONTINUITY

x = a. Some functions have a discontinuity at some
point, but it is possible to define or redefine the
5. their quotient, that is f , if g(a) ≠ 0, is function at that point to make it continuous. These
continuous at x = a. g types of functions are said to have a removable
discontinuity. Let us look at the function f(x)
6. their composite function, f[g(x)] or g[f(x)], represented by the graph in Figure 8.1 or Figure
that is fog(x) or gof(x), is continuous at x = a. 8.3. The function has a limit. However, there is a
hole or gap at x = a. f(x) is not defined at x = a.
8.1.6 TYPES OF DISCONTINUITIES That can be corrected by defining f(x) at x = a.

We have seen that discontinuities have several A function f(x) has a discontinuity at
different types. Let us classify the types of x = a, and lim f (x) exists, but either f(a) is not
discontinuities.
x→a
8.1.7 JUMP DISCONTINUITY
defined or lim f (x) ≠ f (a) . In such case we
As in figure 8.2, for a function, both left-hand x→a
limit and right-hand limits may exist but they
are different. So the graph “jumps” at x = a. The define or redefine f(a) as lim f (x). Then with
function is said to have a jump discontinuity. x→a

A function f(x) has a Jump Discontinuity at new definition, the function f(x) becomes
continuous at x = a. Such a discontinuity is
x = a if the left hand and right-hand limits called a Removable discontinuity.
both exist but are different, that is
If the original function is not defined at a and the
lim f (x) ≠ lim f (x) new definition of f makes it continuous at a, then
xoa xoa the new definition is called the extension of the
original function.

163

Illustration 6: 1
x
Consider f (x) = x2 3x 10 , for x ≠ 2. Observe the graph of xy = 1. y = f(x) = is the
x3 8
function to be considered. It is easy to see that

Here f(2) is not defined. f(x) → ∞ as x → 0+ and f(x) → −∞ as x → 0–.

§ x2 3x 10 · f(0) is not defined. Of course, this function is

lim f(x) = lim ¨ x3 8 ¸ discontinuous at x = 0.
© ¹
x→2 x→2

lim § x 2 x 5 · A function f(x) is said to have an infinite
¨ ¸ discontinuity at x = a,
x→2 ¸¹
=
©¨ x 2 x2 2x 4 if lim f (x) = ±∞ or lim f (x) = ±∞
xoa xoa

= lim § x 5 · = 2+5 = 7 Fig. 8.7 says, f(x) has an infinite discontinuity.
©¨ x2 2x 4 ¸¹ 4+4+4 12
x→2

∴ lim § x2 3x 10 · 7 8.1.10 CONTINUITY OVER AN INTERVAL
¨ ¸ = 12 So far we have explored the concept of continuity
x→2 © x3 8 ¹ of a function at a point. Now we will extend the

idea of continuity on an interval.

Here f(2) is not defined but lim f(x) exists. Let (a, b) be an open interval. If for every
x→2
x ∈ (a, b), f is continuous at x then we say that
Hence f(x) has a removable discontinuity. f is continuous on (a, b) .
The extension of the original function is

x2 3x 10 Consider f defined on [a, b). If f is continuous
f(x) = x3 8 for x ≠ 2 on (a, b) and f is continuous to the right of a,
lim f (x) = f(a) then f is continuous on [a, b)
7
= 12 for x = 2 xoa

This is coninuous at x = 2. Consider f defined on (a, b]. If f is continuous
on (a, b) and f is continuous to the left of b
8.1.9 INFINITE DISCONTINUITY
lim f (x) = f(b), then f is continuous on (a, b]

xoa

Consider a function f continuous on the open

interval (a, b). If lim f (x) and lim f (x) exists,

xoa xob

then we can extend the function to [a, b] so that it

is continuous on [a, b].

SOLVED EXAMPLES

Fig. 8.7 Ex. 1. : Discuss the continuity of the function
f (x) = |x − 3| at x = 3.
Solution : By definition of a modulus function,
the given function can be rewritten as

164

f(x) = – (x – 3) if x < 3 and lim f (x) = 7,
= x – 3 if x ≥ 3 xo2

So lim f (x) = lim f (x) = 7 ⇒ lim f (x) = 7
Now, for x = 3, f(3) = 3 – 3 = 0. xo2 xo2 x→2

lim f (x) = – (3 – 3) = 0 and Also, lim f (x) = f (2) = 7
x→2
xo3

lim f (x) =3 – 3 = 0 ∴ f(x) is continuous at x = 2.

xo3 Let us check the continuity at x = 4.

so, lim f (x) = lim f (x) = 0 ⇒ lim f (x) = 0 f(4) = (42 − 8) = 8
xo3 xo3 x→3

lim f (x) = lim (7) = 7 and
xo4 x→4

lim f (x) = lim (x2 − 8) = 42 – 8 = 8
xo4 x→4

lim f (x) ≠ lim f (x)
xo4 xo4

Fig. 8.8 so lim f (x) does not exist.
x→4
and lim f (x) = f (3) = 0
x→3 Since one of the three conditions does not hold at
x = 4, the function. Hence f (x) is discontinuous at
Therefore the function f(x) is continuous at x = 3. x = 4. Therefore the function f (x) is continuous
Ex. 2 : Determine whether the function f is on it’s domain � , except at x = 4. There exists a
continuous on the set of real numbers jump discontinuity at x = 4.

where f(x) = 3x + 1, for x < 2 \ f is discontinuous at x = 4.

= 7, for 2 ≤ x < 4 Ex. 3 : Test whether the function f (x) is continuous
at x = − 4, where
= x2 −8 for x ≥ 4 .
f (x) = x2 +16x + 48 , for x ≠ −4
If it is discontinuous, state the type of discontinuity. x+4

Solution : The function is defined in three parts, by = 8, for x = −4.
polynomial functions, and all polynomial functions
are continuous on their respective domains. Any Solution : f(−4) = 8 (defined)
discontinuity, if at all it exists, would be at the
points where the definition changes. That is at lim f (x) = lim § x2 16x 48 ·
x = 2 and x = 4. ¨ ¸
Let us check at x = 2. xo 4 xo 4 © x 4 ¹

f (2) = 7 ( Given ) = lim § x 4 x 12 ·
¨ ¸
xo 4 © x 4 ¹

= lim (x+12) ....... [∵ x + 4 ≠ 0]
xo 4

lim f (x) = lim (3x +1) = 3(2) + 1 = 7. = − 4 + 12 = 8
∴ lim f (x) = f (−4) = 8

xo 4

∴ by definition, the function f(x) is continuous at

xo2 xo2 x = −4.

165

Ex. 4 : Discuss the continuity of f (x) = 9 − a2 , Test of continuity of f at x = 0.
on the interval [ −3, 3 ]. For x = 0, f(0) = 0

Solution : The domain of f is [−3, 3]. lim f (x) = lim «¬ x ¼» = lim (−1) = −1 and
xo0
xo0 xo0

[Note, f(x) is defined if 9 − x2 ≥ 0] lim f (x) = lim ¬« x ¼» = lim (0) = 0
xo0
Let x = a be any point in the interval ( -3, 3 ) that xo0 xo0
is a ∈ ( −3, 3 ).
lim f (x) ≠ lim f (x)
xo0 xo0

lim f (x) = lim Therefore f (x) is discontinuous at x = 0.
x→a x→a 9 x2

= 9 − a2 = f (a) Test of continuity of f at x = 1.

∴ for a = 3, f(3) = 0 and for a = −3 , f(−3) = 0 For x = 1, f(1) = 1

Now, lim f(x) = f(3) = 0 lim f (x) = lim ¬« x »¼ = lim (0) = 0 and
xo3 xo1
xo1 xo1
and lim f (x) =f(–3) = 0
xo 3 lim f (x) = lim «¬ x »¼ = lim (1) = 1
xo1
xo1 xo1

Thus f (x) is continuous at every point on (−3, 3) lim f (x) ≠ lim f (x)
and also continuous to the right at x = −3 and to xo1 xo1
the left at x = 3.
Therefore f (x) is discontinuous at x = 1.
Hence, f (x) is continuous on [−3, 3 ].
Hence the function f (x) = «¬x¼» is not continuous
Ex. 5 : Show that the function f (x) = «¬x¼» is not at x = 0, 1 in the interval [−1, 2 ).
continuous at x = 0, 1 in the interval [−1, 2)
Ex. 6 : Discuss the continuity of the following
Solution : f (x) = ¬«x¼» for x ∈ [ −1, 3) function at x = 0, where

that is f (x) = −1 for x ∈ [ −1, 0) f (x) = x2 sin § 1 · , for x ≠ 0
©¨ x ¹¸
f (x) = 0 for x ∈ [ 0, 1)

f (x) = 1 for x ∈ [ 1, 2) = 0, for x = 0.

Solution : The function f (x) is defined for all
x ∈ R.

Let’s check the continuity of f (x) at x = 0.

Given, for x = 0, f(0) = 0.

we know that, −1≤ sin §1· ≤ 1 for any x ≠ 0
¨© x ¹¸

Multiplying throughout by x2 we get

− x2 ≤ x2 sin §1· ≤ x2
©¨ x ¹¸

Fig. 8.9 Taking limit as x → 0 throughout we get,

166

lim (−x2) ≤ lim ª x 2 sin § 1 ·º ≤ lim (x2) since lim § sinT · = 1 = lim § tanT ·
x→0 ¬« ©¨ x ¹¸»¼ x→0  ¨© T ¸¹ ©¨ T ¹¸
x→0 T o0 T o0

0 d lim ª x 2 sin § 1 · º d 0 2
¬« ¨© x ¸¹ ¼» ∴k=3
xo0
Ex. 8 : If f is continuous at x = 1, where

∴ by squeeze theorem we get, f (x) = sin(S x) + a, for x < 1
x 1
lim ª x2 sin § 1 · º =0
¬« ©¨ x ¹¸ ¼» = 2π , for x = 1
x→0

lim f (x) = lim ª x 2 sin § 1 · º =0 1 cos(S x)
x→0 «¬ ¨© x ¹¸ »¼ = S (1 x)2 + b, for x > 1,
x→0 then find the values of a and b.

lim f (x) = f (0) = 0 Solution : Given that f (x) is continuous at x = 1

x→0 ∴ lim f (x) = lim f (x) = f (1) ............ (1)
xo1 xo1
∴ f (x) is continuous at x = 0
Now, lim f (x) = f (1)
Ex. 7 : Find k if f(x) is continuous at x = 0, where xo1
f (x) = xex + tan x , for x ≠ 0
lim § sin(S x) a · = 2p
sin 3x ¨© x 1 ¸¹
xo1
= k , for x = 0
Solution : Given that f(x) is continuous at x = 0, Put x – 1 = t, x = 1 + t as x→1, t → 0
∴ f (0) = lim f (x)
lim § sin S (1 t ) a · = 2p
x→0 ¨© t ¹¸
t→0

k = lim § xex tan x · lim §¨© sin(S S t) a ·¹¸
= x→0 3x ¸ t
§ ¨ sin x· ¹ t→0 = 2p
¨ ©
ex tan ¸
lim ¨ x ¸ ¨§© sin S t a ¸·¹ = 2p
x→0 ¨ ¸ lim t
sin 3x
t→0

©x¹

lim(e x ) lim § tan x · − lim § sin S t · ×π+ lim (a) = 2p
¨© x ¹¸ ¨© S t ¸¹
xo0 xo0 t→0 t→0

= § sin 3x · − (1) π + a = 2π ⇒ a = 3p
©¨ 3 x ¸¹
lim u 3

xo0

1+1 1 2  lim § sinT · = 1
= 1 × 3 = 3 as x → 0, 3x → 0  ©¨ T ¹¸
T o0

From (1), lim f (x) = f (1)
xo1

167

lim § 1 cos(S x) · = 2p Solution :
©¨ S (1 x)2 ¹¸
xo1 b (1) f (x) = x2 − 3x −18
x−6
Put 1− x = θ ∴ x =1− θ , as x → 1, θ → 0
Here f (x) is a rational function, which is continuous
lim § 1 cos(S (1 T )) b · for all real values of x, except for x = 6. Therefore
¨© ST ¸¹ f(6) is not defined.
∴ T o0 2 = 2p

∴ lim § 1 cos(S ST ) b · = 2p Now, lim f (x) = lim § x2 3x 18 ·
©¨ ST 2 ¹¸ ¨ ¸
T o0 x→6 x→6 © ¹

x 6

∴ lim § 1 cosST b · = 2p § (x 6)(x 3) ·
©¨ ST 2 ¸¹ ©¨ x 6 ¹¸
T o0 = lim

x→6

§ 2 sin 2 § ST · ·
¨ ¨© 2 ¹¸ ¸
∴ lim ¨ b¸ = 2p = lim (x +3) [(x−6)≠ 0]
©¨¨ ST 2 ¸¸¹
T o0 x→6

∴ lim f (x) = 9

x→6

ª sin § ST · º2 Here f(6) is not defined but lim f (x) exists.
« ¨© 2 ¸¹
« » ·2 x→6
« » ¸¹
2 lim ST §S + lim (b) = 2p Hence f (x) has a removable discontinuity.
p ©¨ 2
T o0 » T o0

«¬ 2 ¼» (2) g(x) = 3x + 1, for x < 3

2 p2 § sinT · = 2 – 3x, for x ≥ 3
p 4 ¨© T ¸¹
(1) × + b = 2p [ lim = 1] This function is defined by different polynomials
on two intervals. So they are continuous on the
T o0 open intervals (– ∞, 3) and (3, ∞).

∴ p + b = 2p ∴b= 3p
2 2

∴ a = 3p ,b= 3p We examine continuity at x = 3.
2
For x = 3, g(3) = 2 – 3(3) = −7

Ex. 9 : Identify discontinuities for the following lim g (x) = lim (3x + 1) = 10 and
functions as either a jump or a removable xo3 xo3
discontinuity on R.
lim g (x) = lim (2 − 3x) = −7
(1) f (x) = x2 − 3x −18 , xo3 xo3
x−6
lim g (x) ≠ lim g (x) ∴ lim g (x) does not exist.
xo3 xo3 x→3

Hence g is not continuous at x = 3.

(2) g (x) = 3x + 1, for x < 3 The function g (x) has a jump discontinuity at
= 2 − 3x, for x ≥ 3 x = 3.

(3) h (x) = 13 − x2, for x < 5 (3) h (x) = 13 − x2 , for x < 5 ,
= 13 – 5x, for x > 5
= 13 – 5x , for x > 5 ,

168

but h(5) is not defined. Solution : f (p / 2) = log 5 − e
h(x) is continuous at any x < 5 and x > 5
§ 5cos x e§ S x · ·
¨ ©¨ 2 ¹¸ ¸

lim h(x) = lim (13 − x2) = 13 − 25 = −12 Now, lim f (x) = lim ¨ cot x ¸
xo5 xo5 xoS xoS ¨© ¹¸
2 2

and lim h(x) = lim (13 – 5x) = 13 − 25 = −12 Let p − x = t, x = p − t as x → p ,t→0
xo5 xo5 2 2 2

So, lim h(x) = lim h(x) ∴ lim h(x) = −12 § 5cos¨©§ S · ·
xo5 xo5 x→5 ¨ 2 t ¹¸ ¸
¨ et ¸
But for x = 5, f(x) is not defined. lim f (x) = lim
¨ § S · ¸
xoS t→0 ¨© cot ¨© 2 t ¹¸ ¹¸
2
So the function h(x) has a removable discontinuity.

Note : = lim § 5sint et ·

We have proved that lim sin x = 1. t→0 ¨ tan t ¸
x © ¹
x→0

Some standard limits are stated without proof. = lim § 5sint 1 et 1 ·
¨ ¸
ex – 1 t→0 © tan t ¹
x
lim = 1

x→0

ax – 1 = lim § (5sin t 1) (et 1) ·
x ¨© tan t ¸¹
lim = log a t→0

x→0

1

lim (1+ t)t =e § 5sin t 1 et 1 ·
¨ t t ¸
x→0 ¨ tan t ¸
©¨¨ t ¸¸¹
1 = lim

lim (1− t)t = e–1 = 1 t→0
e
x→0

lim log(1 + x) = 1,
x
x→0 [As t→0, t ≠ 0]

lim log(1 – x) = –1 § sin t 5sint 1 et 1 ·
x ¨ t sin t t ¸
x→0 ¨ u tan t ¸
¨¨© t ¸¸¹
These can be proved using L' Hospital's rule, or = lim
expressions in power series which will be studied
t→0

at advanced stage. [sin t ≠ 0]

Ex. 10 : Show that the function ª lim ¨©§ sin T ¸·¹ 1 lim ¨©§ tan T ·¸¹ º
« T T »
T o0 T o0 »
(1).(log 5) −1 «
§ S x · = « § ex 1· § a 1· »
5 ecos x ¨© 2 ¹¸ log a ¼»
p 1 ¬« lim ©¨ ¸¹ 1, lim ©¨ ¹¸
2
f (x) = , for x ≠ xo0 x xo0 x

cot x

= log 5 − e , for x = p lim f (x) = log5 - 1
2
xoS
2

has a removable discontinuity at x = p . f (p / 2) is defined and lim f (x) exists
2 xoS
2

Redefine the function so that it becomes p
2
continuous at x = p . But lim f (x) ≠ f
2 xoS
2

169

∴the function f(x) has removable discontinuity. Ex. 12 :If f(x) is defined on R, discuss the
This discontinuity can be removed by redefining
f (p / 2) = log 5 – 1. continuity of f at x = p , where
2
So the function can be redefined as follows
f (x) 5cos x 5 cos x 2 , for x ≠ p
2
x).log § 2 S 2x ·
(3 cot ©¨ 2 ¸¹

§ 5cos x e§ S x · ·
¨ ©¨ 2 ¸¹ ¸
¸ p
f (x) = ¨ cot x ¸¹ , for x ≠ 2 2 log 5 p
©¨ 2
= 3 , for x = .

= log 5 − 1 , for x = p Solution : Given that, for x = p ,
2 2

1 § S · 2 log 5

Ex. 11 : If f (x) = § 3x 2 ·x , for x ≠ 0, f ©¨ 2 ¹¸ = ,
¨© 2 5x ¸¹ 3

is continuous at x = 0 then find f (0) ªº
« »
lim f (x) = lim « 5cos x 5 cos x 2 »

Solution : Given that f(x) is continuous at x = 0 xoS xoS « 3(cot x) log § 2 S 2x · »
2 ¬« ¨© 2 ¸¹ »¼
2

1

∴ f (0) = lim § 3x 2 ·x ªº
x→0 ©¨ 2 5x ¸¹

= lim « 5cos x 5 cos x 2 »
« »
1 xoS S
2
§ ¨©§1 3x · · x « 3(cot x) log ©¨§1 2x · »
¨ 2 ¸¹ ¸ ¬« 2 ¹¸ ¼»
¨ 2 ¸

= lim ©§¨1 5x ·
2 ¹¸
x→0 ¨ ¸
¨© ¹¸
2 Let S 2x = t ⇒ x = p − t as x → p ,t→0
2 2 2

§ ¨§©1 3x 1 · ∴ lim f (x)
¨ 2 ¸
¨ ·x ¸ xoS
¸¹ 2

= lim ¨ 1¸

x→0 5x § ©§¨ S ·¹¸ ©¨§ S ¹¸· ·
2 ¨ 2 2 ¸
©¨¨ ©§¨1 ·x ¹¸¸ ¨ cos t cos t ¸
¹¸
= lim 5 5 2

3 t→0 ¨ §¨© S ¸¹·.log 1 ¸
©¨¨ 2 ¸¸¹
ª 2 º 2 3 cot t t
«lim »
«¬ xo0 §¨©1 3x ·3x ¼»
2 ¹¸

= 5 § 5sin t 5 sin t 2 ·
¨ ¸
ª 2 º 2 = lim
«lim »
«¬ xo0 ¨©§1 5x ·5x »¼ t→0 ©¨ 3(tan t).log 1 t ¸¹
2 ¹¸

= § t ·
3 ¨ 5sin 5sin t 5 sin t 2 ¸
¸
= e2 ª¬«lxiom0 1 kx 1 e»¼º lim ¨ 3(tan t) log 1 t .5sin t ¸
kx ¨
−5 t→0

e2 ©¹

f(0) = e4 .........[Multiply and divide by 5sin t ]

170

= lim § (5sin t )2 1 2 u 5sin t · Activity 1 :
¨ ¸ Discuss the continuity of f(x)
t→0 ©¨ 3(tan t) log 1 t (5sin t ) ¹¸
log x − log 5
= lim § (5sin t 1)2 · where f(x) = x − 5 for x ≠ 5
¨ ¸
t→0 ¨© 3(tan t).log 1 t .(5sin t ) ¹¸

ª (5sint 1)2 u sin2 t º 1 ........... (I)
« sin2 t » = 5 for x = 5
= lim « »
« » Solution. : Given that f(5) =
t→0 ¬« »¼
3(tan t). log(1 t ).(5sin t )

∴ lim f (x) lim ª log x log 5 º
xo5 ¬« x 5 »¼
t is small but t ≠ 0. Hence sin t ≠ 0 xo5

So we can multiply and divide the numerator by put x – 5 = t ∴ x = 5 + t. As x → 5, t → 0
sin2t

§ 2 · lim f ( x)= lim log( og 5
¨ ¸ 5
¨ § 5sin t 1 · § sin t 2 ¸ t
= lim ¨ ¨ ¸ ©¨ t ¸
©¨ sin t ¸¹ u ·
¹¸

t→0 ¨ 3©¨§ tan t · log 1 t t ¸ t
¨ t ¸¹ ¸
¨ . t .5sin ¸ = ltim log 5

©¹ t

[Dividing Numerator and Denominator by t2 as = ltim log 1 t
t ≠ 0] 5

ª§ ·º 2
«lim ¨ ¸»
5sin t 1 § sin t ·2 t
©¨ t ¹¸
«¬ to0 ©¨ sin t ¹¸¼» u lim

to0 §1 ·
¸
= § tan t · § log(1 t ) · × lim ¨ 5sin t ¹ t
©¨ t ¸¹ ¨© t ¹¸ © 5
t→0

3 lim .lim log 1 1
= ltim t
to0 to0

(log 5)2 × (1) 1 ª § sin T · 1, lim § tan T · 1 º
= 3(1)(1) × 50 ««lTiom0 ©¨ T ¸¹ ©¨ T ¹¸ »
T o0 »
»
« § log(1 x) · § a x 1 · log a¼»» 1ª § log 1 px · º
«lim ©¨ x ¸¹ ¨ ¸ ¸ 1»
« xo0 1, lim = 1u � «¬«lxiom0 ¨ px ¹ ¼»
¬ xo0 ©¨ x ¸¹ ©

lim (log 5)2 ?lim f (x) 1
3 xo0
∴ xoS f (x) = � ........... (II)

2

§S · ∴ from (I) and (II)
©¨ 2 ¸¹
∴ lim f (x) ≠ f lim f (x) f (5)

xoS xo5
2
∴ The function f(x) is continuous at x = 5.
§S ·
∴ f(x) is discontinuous at x = ¨© 2 ¸¹

171

8.1.11 THE INTERMEDIATE VALUE So by intermediate value theorem there has to be
THEOREM FOR CONTINUOUS a point c between 1 and 2 with f (c) = 0.
FUNCTIONS Hence there is a root for the equation
x3 − x −1 = 0 between 1 and 2.
Theorem : If ƒ is a continuous function on a closed
EXERCISE 8.1
interval [a, b], and if y0 is any value between f(a)
and f(b) then y0 = f(c) for some c in [a, b].

1) Examine the continuity of

(i) f (x) = x3 + 2x2 − x − 2 at x = − 2.

(ii) f (x) = sin x, for x ≤ p
4

= cos x, for x > p , at x = p
4 4

(iii) f (x) = x2 − 9 , for x ≠ 3
x−3
Fig. 8.10
= 8 for x = 3
Geometrically, the Intermediate Value Theorem
says that any horizontal line y = y0 crossing the 2) Examine whether the function is continuous
Y-axis between the numbers ƒ(a) and ƒ(b) will at the points indicated against them.
cross the curve y = f (x) at least once over the
interval [a, b]. The proof of the Intermediate Value (i) f (x) = x3 − 2x + 1 , if x ≤ 2
Theorem depends on the completeness property
of the real number system and can be found in = 3x − 2 , if x > 2 , at x = 2.
more advanced texts. The continuity of ƒ on the
interval is essential. If ƒ is discontinuous at even (ii) f (x) = x2 18x 19 , for x ≠ 1
one point of the interval, the conclusion of the x 1
theorem may fail.
= 20 for x = 1 , at x = 1
Illustration 1 : Show that there is a root for the
equation x3 − x −1 = 0 between 1 and 2 (iii) f (x) = x + 2 , for x < 0
tan 3x
Solution : Let f (x) = x3 − x −1 . f(x) is a polynomial
function so, it is continuous everywhere. We say 7
that root of f(x) exists if = 3 , for x ≥ 0 , at x = 0.

f (x) = 0 for some value of x. 3) Find all the points of discontinuities of
f(x) = ¬«x¼» on the interval (−3, 2).
For x = 1, f (1) = 13 − 1− 1 = −1 < 0
4) Discuss the continuity of the function
so, f (1) < 0 f(x) = |2x + 3| , at x = −3/2

For x = 2, f(2) = 23 − 2 − 1 = 5 > 0 so,

f(2) > 0

172

5) Test the continuity of the following functions (iii) f (x) = x2 − 3x − 2 , for x < −3
at the points or interval indicated against = 3 + 8x , for x > −3.
them. (iv) f(x) = 4 + sin x , for x < π
= 3 – cos x for x > π
1

(i) f (x) = x −1 − (x −1)3 , for x ≠ 2
x−2

1 for x = 2 7) Show that following functions have
= 5 , continuous extension to the point where f(x)
is not defined. Also find the extension
at x = 2

x3 8 (i) f (x) = 1− cos 2x , for x ≠ 0.
sin x
(ii) f (x) = for x ≠ 2
x 2 3x 2
3sin2 x 2 cos x(1 cos 2x)
(ii) f (x) = , for x ≠ 0.
= - 24 for x = 2 at x = 2 2 1 cos2 x

(iii) f (x) = 4x + 1, for x ≤ 8/3. (iii) f (x) = x2 1 for x ≠ −1.
59 − 9x x3 1

= 3 , for x > 8/3, at x = 8/3. 8) Discuss the continuity of the following
functions at the points indicated against
1 them.

(iv) f (x) = (27 2x)3 3 1 , for x ≠ 0

9 3(243 5x)5 (i) f (x) = 3 tan x ,x≠ p
S 3x 3

= 2 for x = 0 , at x = 0 3 for x = p , at x = p .
= 4 , 3 3

(v) f (x) = x2 8x 20 for 0 < x < 3; x≠2 (ii) f (x) = e1/ x 1 , for x ≠ 0
2x2 9x 10 e1/ x 1 for x = 0 , at x = 0.

= 12 , for x = 2 = 1 ,

= 2 − 2x − x2 for 3 ≤ x < 4 4x 2x 1 1
x−4 (iii) f (x) = 1 cos 2x , for x ≠ 0

at x = 2

= (log 2)2 , for x = 0 , at x = 0.

6) Identify discontinuities for the following 2
functions as either a jump or a removable
discontinuity. 9) Which of the following functions has
a removable discontinuity? If it has a
(i) f (x) = x2 10x 21 . removable discontinuity, redefine the
x 7 function so that it becomes continuous.

(ii) f (x) = x2 + 3x − 2 , for x ≤ 4 (i) f (x) = e5sin x − e2x , for x ≠ 0
= 5x + 3 , for x > 4. 5 tan x − 3x

= 3/4 , for x = 0 , at x = 0.

173

(ii) f (x) = log(1+3x) (1+5x) for x > 0 (iii) If f (x) = sin 2x − a , for x > 0
32x −1 5x

= 8x −1 , for x < 0 , at x = 0. = 4 for x = 0

= x2 + b − 3 , for x < 0

1 is continuous at x = 0, find a and b.

(iii) f (x) = § 3 8x ·x , for x ≠ 0.
©¨ 3 2x ¸¹
(iv) For what values of a and b is the function
(iv) f (x) = 3x + 2 , for −4 ≤ x ≤ −2 f (x) = ax + 2b + 18 , for x ≤ 0
= 2x – 3 , for −2< x ≤ 6. = x2 + 3a − b , for 0 < x ≤ 2
= 8x – 2 , for x > 2,
x3 − 8 continuous for every x ?
x2 − 4
(v) f (x) = , for x > 2

= 3 , for x = 2 (v) For what values of a and b is the function

= e3( x− 2)2 −1 , for x < 2 f (x) = x2 − 4 , for x < 2
2( − 2)2 x−2
x

= ax2 − bx + 3 , for 2 ≤ x < 3

2 sin x 3 p = 2x – a + b , for x ≥ 3
cos2 x , for x ≠ 2
10) (i) If f (x) = , continuous for every x on R?

p §S · 12) Discuss the continuity of f on its domain,
2 ©¨ 2 ¹¸ .
is continuous at x = then find f where

(ii) If f (x) = cos2 x sin2 x 1 for x ≠ 0, f (x) = |x + 1| , for −3 ≤ x ≤ 2
3x2 1 1
= |x − 5| , for 2 < x ≤ 7 .
is continuous at x = 0 then find f(0).
13) Discuss the continuity of f(x) at x = p
4x S 4S x 2 where, 4
(iii) If f (x) = (x S )2 for x ≠ p ,
is continuous at x = p, then find f(p) . f (x) = (sin x cos x)3 2 2 , for x ≠ p
sin 2x 1 4

11) (i) If f (x) = 24x 8x 3x 1 , for x ≠ 0 = 3 for x = p .
12x 4x 3x 1 , 4

2

= k , for x = 0 14) Determine the values of p and q such that
the following function is continuous on the
is continuous at x = 0, find k. entire real number line.

5x 5 x 2 f (x) = x + 1 , for 1 < x < 3
(ii) If f (x) = x2 , for x ≠ 0 = x2 + px + q , for |x − 2| ≥ 1.
= k for x = 0

is continuous at x = 0, find k.

174

15) Show that there is a root for the equation Let's Remember
2x3 − x − 16 = 0 between 2 and 3.
Continuity at a point
16) Show that there is a root for the equation
x3 − 3x = 0 between 1and 2. A function f (x) is continuous at a point a if and
only if the following three conditions are satisfied:
17) Activity : Let f(x) = ax + b (where a and b
are unknown) (1) f (a) is defined, (2) lim f (x) exists, and
x→a
= x2 + 5 for x ∈ R
(3) lim f (x) = f (a)
Find the values of a and b, so that f(x) is x→a
continuous at x = 1. (Fig. 8.11)
Continuity from right : A function is continuous

from right at a if lim f (x) = f (a)

xoa

Continuity from left : A function is continuous
from left at b if lim f (x) = f (b)

xob

Continuity over an interval :

Open Interval : A function is continuous over an
open interval if it is continuous at every point in
the interval.

Fig. 8.11 Closed Interval : A function f (x) is continuous
over a closed interval [a,b] if it is continuous at
18) Activity : Suppose f(x) = px + 3 for a ≤ x ≤ b every point in (a,b),and it is continuous from
= 5x2 − q for b < x ≤ c right at a and from left at b.
Find the condition on p, q, so that f(x) is
Discontinuity at a point :
continuous on [a,c], by filling in the boxes. A function is discontinuous at a point or has a pont
f(b) = of discontinuity if it is not continuous at that point

Infinite discontinuity :
An infinite discontinuity occurs at a point a if

lim f (x) = ±∞ or lim f (x) = ±∞
lim f(x) = xoa xoa
xob
–q Jump discontinuity :
∴ pb + 3 =
A jump discontinuity occurs at a point ‘a’ if

∴ p = b is the required condition. lim f (x) and lim f (x) both exist, but
xoa xoa

lim f (x) ≠ lim f (x)
xoa xoa

175

Removable discontinuity : (4) f (x) = x2 7x 10 , for x ∈ [−6, −3]
A removable discontinuity occurs at a point a if x2 2x 8
f (x) is discontinuous at a, but lim f (x) exists and
(A) f is discontinuous at x = 2.
x→a (B) f is discontinuous at x = -4.
(C) f is discontinuous at x = 0.
f (a) may or may not be defined. (D) f is discontinuous at x = 2 and x = -4.

Intermediate Value Theorem : (5) If f (x) = ax2 + bx + 1, for |x −1| ≥ 3 and
Let f be continuous over a closed bounded interval
[a,b]. If z is any real number between = 4x + 5, for -2 < x < 4

f (a) and f (b), then there is a number c in [a,b] is continuous everywhere then,
satisfying f (c) = z.

MISCELLANEOUS EXERCISE-8 11
(A) a = , b = 3 (B) a = - , b = - 3

22

(C) a = - 1 , b = 3 1
(D) a = , b = -3
(I) Select the correct answer from the given 22
alternatives.

2cot x 1 for x ≠ p 16x 1 9x 1
(1) f (x) = S 2x , 2 (6) f (x) = 27x 1 32x 1 , for x ≠ 0
= log 2 ,
for x = p = k , for x = 0
2 is continuous at x = 0, then ‘k’ =

(A) f is continuous at x = p
2
p 88 8 20
(B) f has a jump discontinuity at x = 2 (A) (B) (C) - (D)

(C) f has a removable discontinuity 3 15 15 3

(D) lim f (x) = 2log3 32x 8x 4x 1
(7) f (x) = 4x 2x 1 1 , for x ≠ 0
xoS
2 = k , for x = 0,
is continuous at x = 0, then value of ‘k’ is
(2) If f (x) = 1 2 sin x , for x ≠ p
S 4x 4 (A) 6 (B) 4 (C) (log2)(log4) (D) 3log4

is continuous at x = p , then f § S · =
4 ©¨ 4 ¸¹

11 1 1 (8) If f (x) = 12x 4x 3x 1 , for x ≠ 0 is
(A) (B) - (C) - (D) 1 cos 2x
2 24
4 continuous at x = 0 then the value of f(0) is

sin 2x tan 5x (A) log12 (B) log2.log3

(3) If f (x) = (e2x 1)2 , for x ≠ 0 2
is continuous at x = 0, then f(0) is
(C) log 2.log 3 (D) None of these.

10 10 5 5 2
(A) e2 (B) e4 4 (D)
(C)
2

176

4 x 1
x2 x
(9) If f (x) = § 4 5x · x , for x ≠ 0 and f(0) = k, is (5) f (x) = , for x ≠ −1
©¨ 4 7x ¸¹
2 1

continuous at x = 0, then k is = 0 for x = -1 at x = -1.

(A) e7 (B) e3 (C) e12 3 (6) f (x) = > x 1@ for x ∈ [−2, 2)

(D) e4 Where [*] is greatest integer function.

(10) If f(x) = «¬x¼» for x∈(−1,2) then f is (7) f (x) = 2x2 + x + 1, for |x − 3| ≥ 2
discontinuous at = x2+3 , for 1 < x < 5

(A) x = -1, 0, 1, 2, (B) x = -1, 0, 1

(C) x = 0, 1 (D) x = 2 (III) Identify discontinuities if any for the
following functions as either a jump or a
(II) Discuss the continuity of the following removable discontinuity on their respective
functions at the point(s) or on the interval domains.
indicated against them.

(1) f (x) = x2 − 3x −10 , for 3 ≤ x ≤ 6, x ≠ 5 (1) f (x) = x2 + x - 3 , for x ∈ [ -5, -2 )
x−5 = x2 - 5 , for x ∈ ( -2, 5 ]

= 10 , for x = 5 (2) f (x) = x2 + 5x + 1 , for 0 ≤ x ≤ 3
= x3 + x + 5 , for 3 < x ≤ 6
= x2 − 3x −10 , for 6 < x ≤ 9
x−5
(3) f (x) = x2 + x +1 , for x ∈ [ 0, 3 )
(2) f (x) = 2x2 −2x + 5 , for 0 ≤ x ≤ 2 x +1

= 1− 3x − x2 , for 2 < x < 4 = 3x 4 , for x ∈ [ 3, 6].
1− x x2 5

x2 − 25 (IV) Discuss the continuity of the following
functions at the point or on the interval
= x − 5 , for 4 ≤ x ≤ 7 and x ≠ 5 indicated against them. If the function
is discontinuous, identify the type of
= 7 for x = 5 discontinuity and state whether the
discontinuity is removable. If it has a
cos 4x − cos 9x removable discontinuity, redefine the
(3) f (x) = 1− cos x , for x ≠ 0 function so that it becomes continuous.

f (0) = 68 = 0 on S d xdS (x 3)(x2 6x 8)
, at x
15 2 2
(1) f (x) =
x2 x 12

sin2 S x (2) f (x) = x2 + 2x + 5, for x ≤ 3
(4) f (x) = 3(1 x)2 , for x ≠ 1
= x3 − 2x2 − 5, for x > 3

S 2 sin 2 § Sx ·
¨© 2 ¸¹

= 2 § Sx · for x = 1, at x = 1.
¨© 2 ¹¸
3 4 cos

177

(V) Find k if following functions are continuous (2) f (x) = ax2 + bx + 1 , for |2x − 3| ≥ 2
at the points indicated against them. 15

3 = 3x + 2 , for 2 < x < 2 .

(1) f (x) = § 5x 8 ·2x 4 , for x ≠ 2 (VII) Find f(a), if f is continuous at x = a where,
¨© 8 3x ¸¹

= k , for x = 2 at x = 2. 1 cos(S x)
(1) f (x) = S (1 x)2 , for x ≠ 1 and
45x 9x 5x 1 at a = 1.
(2) f (x) = (k x 1)(3x 1) , for x ≠ 0

2 for x = 0, at x = 0 1 cos[7(x S )]
= 3 , (2) f (x) = 5(x S )2 , for x ≠ p at

(VI) Find a and b if following functions are a = p.
continuous at the points or on the interval
indicated against them. (VIII) Solve using intermediate value theorem.

(1) f (x) = 4 tan x 5sin x , for x < 0 (1) Show that 5x − 6x = 0 has a root in [1, 2]
ax 1
(2) Show that x3 − 5x2 + 3x + 6 = 0 has at
9 for x = 0 least two real roots between x = 1 and
= log 2 , x = 5.

11x 7x.cos x
= bx 1 , for x > 0.

vvv

178

9 DIFFERENTIATION

Let's :Learn When we speak of velocity, it is the speed with
the direction of movements. In problems with
• The meaning of rate of change. no change in direction, words speed and velocity
• Definition of derivative and the formula may be interchanged.

associated with it. The rate of change in a function at a point with
• Derivatives of some standard functions. respect to the variable is called the derivative of
• Relationship between Continuity and the function at that point. The process of finding a
derivative is called differentiation
Differentiability.
9.1.2 DEFINITION OF DERIVATIVE AND
Let's Recall DIFFERENTIABILITY

• Real valued functions on R. Let f(x) be a function defined on an open interval
• Limits of functions. containing the point ‘a’. If
• Continuity of a function at a point and over
lim § f (a G x) f (a) · exists, then f is said to
an interval. ¨© Gx ¹¸
G xo0
9.1.1 INTRODUCTION :
be differentiable at x = a and this limit is said to
Suppose we are travelling in a car from be the derivative of f at a and is denoted by f ′(a).
Mumbai to Pune. We are displacing ourselves
from the origin (Mumbai) from time to time. We We can calculate derivative of ‘f  ’ at any point x in
know that the average speed of the car the domain of f.

Total distance travelled Let y = f (x) be a function. Let there be a small
= Time taken to travel that distance increment in the value of ‘x’, say δ x , then
correspondingly there will be a small increment
But at different times the speed of the car can be in the value of y say δ y .
different. It is the ratio of a very small distance
travelled, with the small time interval required to ? y G y f (x G x) .... [  y f (x)]
travel that distance. The limit of this ratio as the ? G y f (x G x) y
time interval tends to zero is the speed of the car
at that time. This process of obtaining the speed G y f (x G x) f (x)
is given by the differentiation of the distance
function with respect to time. This is an example As δ x is a small increment and δx ≠ 0, so dividing
of derivative or differentiation. This measures
how quickly the car moves with time. Speed is throughout by δ x , we get G y f (x G x) f (x)
the rate of change of distance with time. G x Gx

Now, taking the limit as δ x → 0 we get

lim § G y · lim § f (x G x) f (x) ·
¨© G x ¸¹ ¨© Gx ¸¹
G xo0 G xo0

179

If the above limit exists, then that limiting value f '(x) lim § f (x h) f (x) · = dy = f '( x)
is called the derivative of the function and it is ¨© h ¸¹ dx
symbolically represented as, dy = f '(x) ho0

dx The derivative of y = f(x) with respect to x at x = a
by method of first principle is given by
so dy = f '(x)
dx § f (a h) f (a) · = ©¨§ dy ·
©¨ h ¹¸ dx ¹¸ x
We can consider the graph of f(x) i.e. f c(a) lim a
{(x, y) / y = f(x)} and write the differentiation in
terms of y and x ho0

NOTE : (1) If y = f(x) is a differentiable function 9.1.4 DERIVATIVESOFSOMESTANDARD
FUNCTIONS
of x then lim § G y · dy and
©¨ G x ¸¹ dx (1) Find the derivative of xn w. r. t. x. for n∈N
G xo0
Solution :
lim § f (x G x) f (x) · f '(x)
©¨ Gx ¹¸ Let f (x) = xn
G xo0

(2) Let δ x = h , Suppose that f (x + h) = (x + h)n
By method of first principle,
lim § f (a h) f (a) · exists. It implies that
¨© h ¸¹
ho0 f (x h) f (x) ·
h ¹¸
§ f (a h) f (a) · § f (a h) f (a) · f c(x) lim §
¨© h ¸¹ ¨© h ¹¸ ¨©
= ho0

lim lim h)n xn
h
ho0 ho0

f c(x) lim § (x ·
¨ ¸
lim § f (a h) f (a) · is called the Left Hand ho0 © ¹
©¨ h ¹¸ Derivative (LHD)at x =
ho0 a

= lim § xn nC1xn 1h nC2 xn 2h2 ....... hn xn ·
h ¸
 f (a + h) − f (a)  ho0 ¨ ¹
 h  ©
lim is called the Right Hand
Derivative (RHD)at x = a § nC1 x n 1h nC2 xn 2h2 nC3 x hn 3 3 ..... hn ·
h→0+ h ¸
¹
= lim ¨
©
Generally LHD at x = a is represented as f ′(a-) ho0
or L f ′(a), and RHD at x = a is represented as
f ′(a+) or R f ′(a) = lim § h( nC1xn 1 nC2 xn 2h nC3 x hn 3 2 ..... h n 1 ) ·
¨ h ¸
ho0 © ¹

=
9.1.3 DERIVATIVE BYMETHOD OFFIRST lim nxn 1 nC2 xn 2h nC3x hn 3 2 ..... hn 1
PRINCIPLE.
ho0
The process of finding the derivative of a function
from the definition of derivative is known as ... (as h →0 , h ≠ 0)
derivatives from first principle. Just for the sake
of convenience δx can be replaced by h. ∴ if f(x) = xn, f (x) = nxn−1

If f(x) is a given function on an open interval, then
the derivative of f(x) with respect to x by method
of first principle is given by

180

(2) Find derivative of sin x w. r. t. x. f c(x) lim § tan( x h) tan x ·
Solution : ¨© h ¸¹
ho0

Let f (x) = sin x § sin(x h) sin x ·
¨ cos(x h) cos x ¸
= lim ¨ ¸
f (x + h=) sin(x + h) ho 0 ¨¨© h ¸¸¹

By method of first principle, § sin(x h).cos x cos(x h) sin x ·

f c(x) lim § f (x h) f (x) · = lim ¨¨ cos(x h).cos x ¸
©¨ h ¸¹ ¸
ho0 ho0 ¨¨© h ¸¸¹

f c(x) lim § sin( x h) sin x ·
©¨ h ¸¹
ho0 § sin h ·
¨ x h). ¸
= lim © h.cos( cos x ¹

§ 2x h h · ho0
¨ 2 2 ¸
¨ 2cos § ¹·¸.sin § · ¸
¨© ¨© ¸¹
lim § sin h · § 1 ·
©¨ h ¹¸ ¨ h). ¸
ho0 ¨ h ¸ = lim . lim © ¹

©¨ ¹¸ ho0 ho0 cos( x cos x

2x h § sin § h · · = (1).©§¨ 1 · ........ ª«¬Tliom0 § sin T · º
2 ¨ ©¨ 2 ¸¹ ¸ cos2 ¹¸ ¨© T ¹¸ 1¼»
= 2 lim cos § · lim ¨ h ¸ § 1 · x
¨© ¹¸ ¨ ¸ ¨© 2 ¹¸
ho0 ho0

©¨ 2 ¸¹ ∴ if f(x) = tan x, f (x) = sec2x

§ sin § h · ·
¨ ©¨ 2 ¹¸ ¸ ( 4) F ind the derivative of sec x w. r. t. x.
= 2 lim cos § x h · lim ¨ ¸ § 1 ·
¨© 2 ¹¸ ¨ h ¸ ¨© 2 ¸¹ Solution:
ho0 ho0
Let f (x) = sec x
©¨ 2 ¹¸

= 2 cos x.(1) § 1 · ....... ª lim § sin pT · º f (x + h=) sec(x + h)
¨© 2 ¹¸ « ¨ pT ¸ 1» From the definition,
¬ T o0 © ¹ ¼

∴ if f(x) = sin x, f (x) = cosx

f c(x) lim § f (x h) f (x) ·
¨© h ¸¹
(3) Find the derivative of tan x w. r. t. x. ho0
Solution:
Let f (x) = tan x
f (x + h=) tan(x + h)
f c(x) lim § sec( x h) sec x ·
From the definition, ©¨ h ¹¸
ho0



§ 1 h) 1 x ·
¨ cos(x cos ¸
lim ¨ ¸

f c(x) lim § f (x h) f (x) · ho0 ¨©¨ h ¸¹¸
©¨ h ¹¸
ho0

181

§ cos x cos(x h) · § log ¨©§1 h · ·
¨ ¸ ¨ h x ¹¸ ¸
= lim ¨ cos(x h).cos x ¸ = lim ¨ ¸ u § 1 ·
¨ ¸ ¨© x ¹¸
ho0 ¨¨© h ¸¹¸ ho0

©¨ x ¸¹

§ 2 sin § 2x h · .sin § h · ·
¨ ©¨ 2 ¸¹ ©¨ 2 ¸¹ ¸
= lim ¨ ¸ = 1. § 1 ·  lim  log(1 + x)  = 1
¨© x ¸¹  x 
ho0 ¨ h.cos(x h).cos x ¸ x→0

¨© ¸¹

= 2 hliom 0 ¨¨¨§ cossin(x©¨§ 2hx2) .cho·¹¸s x ¸¸·¸ hliom0 §¨¨¨ s i nh§©¨ h2 ·¸¹ ¸·¸¸ ¨§© 12 ¸¹· ∴ if f(x) = log x, f (x) = 1
x

¨© ¹¸ ¨© 2 ¹¸ (6) Find then derivative of w. r. t. x. (a > 0)

= 2 sin x .(1).©¨§ 1 · ....... ª«¬Tliom0 § sin pT · º Solution:
cos2 x 2 ¸¹ ¨ pT ¸ 1» Let f (x) = ax
© ¹ ¼

∴ if f(x) = sec x, f (x) = secx , tanx f (x + h) =ax+h
From the definition,
(5) Find the derivative of log x w. r. t. x.(x > 0)
Solution: f c(x) lim § f (x h) f (x) ·
©¨ h ¹¸
Let f (x) = log x ho0
f (x + h=) log(x + h)
f c(x) lim § a(x h) ax ·
From the definition, ©¨¨ h ¸¹¸
ho0

f c(x) lim § f (x h) f (x) · § a x (ah 1) ·
©¨ h ¹¸ ¨ h ¸
ho0 = lim © ¹

ho0



f c(x) lim § log( x h) log( x) · § ah 1 ·
¨© h ¸¹ ¨ h ¸
ho0 = ax lim © ¹

ho0

§ log § x h · · § log ¨§©1 h · · x ª § a x 1 · º
¨ ¨© x ¸¹ ¸ ¨ x ¹¸ ¸ « ¨ x ¸ log a»
lim ¨ ¸ ¨ ¸ f c(x) a log a......... ¬ lim © ¹
ho0 ¨ h ¸ ¼
= lim xo0

¨© ¹¸ ho0 ¨ h ¸
©¨ ¹¸
∴ if f(x) = ax, f (x) = ax.loga

§ log § x h·· § log ©§¨1 h··
¨ ¨© ¹¸ ¸ ¨ ¸¹ ¸
lim ¨ x ¸¸= lim ¨ x ¸
ho0 ¨ h
©¨ ho0 ¨ h ¸
¹¸ ¨© ¹¸

182

Try the following § h ·
x h
(1) If 1 , for x ≠ 0, n∈N, then prove that = lim ¨ h( x ) ¸
xn © ¹
f (x) = ho0

f c(x) n = lim § 1 · ...[As h0, h ≠ 0]
xn 1 ¨© x h ¸¹
ho0 x

(2) If f(x) = cos x, then prove that 1
f ′(x) = − sin x =

2x

(3) If f(x) = cot x, then prove that (ii) Let f(x) = cos (2x + 3)
f ′(x) = − cos ec2x f(x+h) = cos (2(x+h)+3) =cos ((2x+3)+2h)
From the definition,
(4) If f(x) = cosec x, then prove that
f ′(x) = − cos ecx.cot x f c(x) lim § f (x h) f (x) ·
¨© h ¸¹
(5) If f (x) = ex , then prove that f ′(x) = ex ho0



SOLVED EXAMPLES

Ex. 1. Find the derivatives of the following from f c(x) lim § cos[(2x 3) 2h] cos(2 x 3) ·
the definition, ¨© h ¹¸
ho0
(i) x (ii) cos (2x+3) (iii) 4x (iv) log(3x−2)
Solution : § 2 sin § 2(2x 3) 2h · sin(h) ·
(i) Let f (x) = x ¨ ©¨ 2 ¹¸ ¸
= lim ¨ ¸
f (x + h) = x + h
From the definition, ho0 ¨ h ¸
¨© ¸¹

lim § 2 sin(2 x 3 h) sin(h) ·
©¨ h ¸¹
= ho0

== lim[−2 sin(2 x + 3 + h)] lim  sin h 
 h 
h→0 h→0

f c(x) lim § f (x h) f (x) · = «ª¬ § sin T · º
©¨ h ¹¸ ©¨ T ¸¹ 1»¼
ho0 [ 2 sin(2 x 3)](1)....... lim

T o0

f c(x) lim § x h x·
¨©¨ h ¸¸¹
ho0 f c(=x ) 2sin(2x 3)

lim § x h xu x h x· (iii) Let f (x) = 4x
©¨¨ h x h x ¸¸¹
= ho0 f (x + h) =4x+h
From the definition
= lim § x h x ·
¨ x h ¸
ho0 © h( x ) ¹

183

f c(x) lim § f (x h) f (x) · u § 3 · ª § log(1 px) · º
¨© h ¸¹ = (1) ¨© 3x ¹¸ « lim ¨ px ¸ 1»
ho0 ¬ © ¹ ¼
2 xo0

f c(x) lim § 4x h 4x ·
¨ ¸
ho0 © h ¹ f c(x) 3
3x 2
§ 4x (4h 1) ·
= lim ¨ ¸
© h ¹
ho0 Ex. 2. Find the derivative of f(x) = sin x, at
x=π
= 4x lim § 4h 1·
¨ ¸ Solution:
ho0 © h ¹

f c(x) 4x log 4...... ª lim § a x 1 · º f=(x) s=in x si=n π 0
« ¨ x ¸ log a»
¬ xo0 © ¹ f(π) = sin π = 0
¼
f (S h) sin(S h) sin h
From the definition,

(iv) Let =f (x) log(3x − 2) f c(a) lim § f (a h) f (a) ·
f (x +=h) log[3(x + h) −=2] log[(3x − 2) + 3h] ¨© h ¹¸
ho0

From the definition, f c(S ) lim § f (S h) f (S ) ·
©¨ h ¹¸
§ f (x h) f (x) · ho0
©¨ h ¸¹
f c(x) lim

ho0

= lim § sin h 0 · lim § sin h ·
¨© h ¸¹ ¨© h ¹¸
§ log[(3x 2) 3h] log(3x 2) · ho0 ho0
¨© h ¹¸
f c(x) lim

ho0 § sin h ·
¨© h ¹¸
= lim

ho0

§ log § (3x 2) 3h · ·
¨ ©¨ 3x 2 ¹¸ ¸
lim ¨ ¸ –1 1ª¬«Tliom0 § sin T · º
= = ©¨ T ¸¹ 1»¼
ho0 ¨ h ¸
¨© ¹¸
Ex. 3. Find the derivative of x2 + x + 2 , at
== lim  log 1+ 3h 2   x=–3
 3x −  
  Solution :
h→0  h 
 
Let f (x) = x2 + x + 2

 log 1 + 3h 2    3  For x = – 3, f(– 3) = (–3)2 – 3 = 9 – 3 + 2 = 8
 3x −    3x − 
== lim   × f ( 3 h) ( 3 h)2 ( 3 h) 2
 3h  2 h2 6h 9 3 h 2 h2 5h 8
h→0 From the definition,

 3x − 2 

184

f c(a) lim § f (a h) f (a) · lim ª h § f (a h) f (a) ·º lim>hf '(a)@
¨© h ¹¸ ¬« ¨© h ¹¸»¼
ho0 ho0 ho0

f c( 3) lim § f ( 3 h) f ( 3) · lim [f(a + h) − f(a)] = 0[f’(a)] = 0
©¨ h ¹¸
ho0 h→0

∴ lim f(a + h) = f(a)

h2 h→0

lim § 5h 8 8 ·
h ¸
= ho0 ¨ ¹ This proves that f(x) is continuous at x = a.
©
Hence every differentiable function is continuous.

= lim § h2 5h · Note that a continuous function need not be
¨ h ¸ differentiable.
ho0 © ¹

= lim § h(h 5) · This can be proved by an example.
©¨ h ¹¸
ho0 Ex.: Let f (x) = x be defined on R.

= lim (h 5)......[h o 0, h z 0] f(x) = - x for x < 0

ho0

f c( 3) 5 = x for x ≥ 0

Consider, lim f (x) lim ( x) 0

9.1.5 RELATIONSHIP BETWEEN xo0 xo0
DIFFERENTIABILITY AND
CONTINUITY lim f (x) lim ( x) 0

xo0 xo0

Theorem : Every differentiable function is For, x = 0, f(0) = 0
continuous.
lim f (x) lim ( x) f (0) 0

Proof : Let f(x) be differentiable at x = a. xo0 xo0

Then, f c(a) lim § f (a h) f (a) · ........... (1) Hence f(x) is continuous at x = 0.
¨© h ¸¹
ho0 Now, we have to prove that f(x) is not differentiable
at x = 0 i.e. f ′(0) doesn’t exist.
we have to prove that f(x) is continuous at x = a.
i.e. we have to prove that,

i.e. we have to prove that lim f (x) f (a) § f (0 h) f (0) · z § f (0 h) f (0) ·
©¨ h ¹¸ ©¨ h ¹¸
xoa lim lim

Let x = a + h, x→a, h→0 ho0 ho0

We need to show that We have, L. H. D. at x = 0, is f ′(0−)

lim f(a + h) = f(a) = lim § f (0 h) f (0) ·
¨© h ¸¹
h→0 ho0

The equation (1) can also be written as

lim § f (a h) f (a) · lim f '(a) = lim § h · lim ( 1)
¨© h ¹¸ ©¨ h ¹¸
ho0 ho0 ho0 ho0

As h→0, h ≠ 0 f ′(0−) = −1 ……….. (I)

Multiplying both the sides of above equation by Now, R. H. D. at x = 0, is f ′(0+)
h we get

185

= lim § f (0 h) f (0) · § 2· 2 § 2 ·
©¨ h ¸¹ (3h) 5 ¸
ho0 = lim ¨ h ¸ 35 lim ¨ h5 ¸¸¹
¨©¨ ¹¸¸ ¨¨© h
ho0 ho0

= lim § h · lim (1) § ·
¨© h ¹¸ lim ¨ ¸
ho0 ho0 1

= 2

f ′(0+) = 1……….. (II) 35 ho0 ¨ 3 ¸
© h5 ¹

Therefore from (I) and (II), we get This limit does not exist.

f c(0 ) z f c(0 ) that is ∴ f(x) is not differentiable at x = 2
3
§ f (0 h) f (0) · § f (0 h) f (0) ·
lim ¨© h ¸¹ z lim ©¨ h ¹¸
Ex. 2. Examine the differentiability of
ho0 ho0 f (x) =(x − 2) x − 2 at x = 2
Solution : Given that f (x) (x 2) x 2
Though f(x) is continuous at x = 0, it is not That is f (x) (x 2)2 for x < 2
differentiable at x = 0. = (x − 2)2 for x ≥ 2

SOLVED EXAMPLES

2 § f (2 h) f (2) ·
©¨ h ¸¹
Ex. 1. Test whether the function f (x) (3x 2)5

2 Lf c(2) lim
is differentiable at x =
ho0
3
= lim § (2 h 2)2 (2 2)2 ·
Solution : ¨ h ¸
ho0 © ¹
2

Given that, f (=x) (3x − 2)5

f (a h) f (a) · § h 2 ·
h ¹¸ ¨ h ¸
f c(a) lim § = lim © ¹ lim h
¨©
ho0 ho0 ho0


Lf ′(2) = 0
§ 2·
Note that f ©¨ 3 ¹¸ = 0 Rf ′(2) = lim § f (2 h) f (2) ·
h ¸
§ §©¨ 2 h ¹·¸ ©§¨ 2 ¸¹· · ho0 ¨ ¹
¨ 3 3 ¸ ©
lim ¨ f f ¸
2 c ©§¨ 2 · § h 2)2 2)2 ·
For, x , f 3 ¸¹ lim ¨ (2 h (2 ¸
© ¹
3 ho0 ©¨¨ h ¸¸¹ = ho0

§ ª¬«3©¨§ 2 h ¹¸· 2 · = lim § h2 · lim h
¨ 3 h ¸ ¨ h ¸
2¼º» 5 ¸ ho0 © ¹ ho0
¸
= lim ¨ ¸
¨
ho0 ¨ Rf ′(2) = 0
So, Lf ′(2) = Rf ′(2) = 0
©¨ ¹¸

§ >2 3h 2 · Hence the function f(x) is differentiable at x = 2.
¨ ¸
= lim 2@5

ho0 ¨¨© h ¹¸¸

186

Ex. 3. Show the function f (x) is continuous at ∴ f(x) is not differentiable at x = 3.
x = 3, but not differentiable at x = 3. if
Hence f(x) is continuous at x = 3, but not
f(x) = 2x + 1 for x ≤ 3 differentiable at x = 3.

= 16 − x2 for x > 3. Ex. 4. Show that the function f(x) is differentiable
Solution : f(x) = 2x + 1 for x ≤ 3
at x = − 3 where, f (x=) x2 + 2 .
= 16 − x2 for x > 3.
For x = 3 , f(3) = 2(3) + 1 = 7 Solution : f c( 3) lim § f ( 3 h) f ( 3) ·
For x = −3, ¨© h ¹¸
ho0

lim f (x) lim(2x 1) 2(3) 1 7 = lim § ( 3 h)2 2 11 ·
¨ h ¸
xo3 xo3 ho0 © ¹

lim f (x) lim (16 x2 ) 16 (3)2 7 § 9 6h h2 2 11 ·
xo3 xo3 ¨ h ¸
= lim © ¹

lim f (x) lim f (x) f (3) 7 ho0

xo3 xo3

f(x) is continuous at x = 3. = lhiom 0 §¨© h2 6h ·
h ¸
¹

Lf c(3) lim § f (3 h) f (3) · lim h 6 (h→0, h ≠ 0)
©¨ h ¹¸
ho0 = ho0

§ 2(3 h) 1 7 · § 6 2h 1 7 · f c( 3) 6
¨© h ¹¸ ©¨ h ¹¸
= lim lim f c( 3) exists so, f(x) is differentiable at x = − 3.

ho0 ho0

3 h) 1 7 · lim § 6 2h 1 7 ·
h ¸¹ ©¨ h ¹¸
ho0 EXERCISE 9.1

= lim § 2h · lim (2) (1) Find the derivatives of the following w. r. t. x
©¨ h ¹¸ by using method of first principle.
ho0 ho0
(a) x2 + 3x −1 (b) sin (3x)
Lf ′(3) = 2 ……………… (1) (c) e2x+1 (d) 3x (e) log(2x + 5)
(f) tan (2x +3) (g) sec ( 5x − 2)
Rf c(3) lim § f (3 h) f (3) ·
©¨ h ¹¸ (h) x x
ho0

§ 16 (3 h) 2 7· § 16 9 6h h2 7 ·
¨ h ¸ h ¸
= lim © ¹ lim ¨ ¹ (2) Find the derivatives of the following w. r. t. x.
© at the points indicated against them by using
ho0 ho0 method of first principle

= lim § h(6 h) · lim (6 h)
©¨ h ¸¹
ho0 ho0 (a) 2x + 5 at x = 2 (b) tan x at x = π / 4

Rf ′(3) = − 6 ……………… (2) (c) 23x+1 at x = 2 (d) log(2x +1) at x = 2

from (1) and (2), Lf ′(3) ≠ Rf ′(3) (e) e3x − 4 at x = 2 (f) cosx at x = 5π
4

187

(3) Show that the function f is not differentiable Let there be a small increment in the value of
at x = −3,
where f(x) = x2 + 2 for x < − 3 x ,say δ x ,then u changes to (u + δ u) and v
changes to (v + δ v) respectively, correspondingly
= 2 − 3x for x ≥ −3 y changes to ( y + δ y)
(4) Show that f (x) = x2 is continuous and ? y G y u Gu v Gv
? G y u Gu v Gv - y
differentiable at x = 0. ? G y u Gu v Gv - u v

(5) Discuss the continuity and differentiability of > y u v@

(i) f (x) = x x at x = 0 ? G y Gu Gv

(ii) f (x) =(2x + 3) 2x + 3 at x = − 3/2 As δ x is small increment in x and δ x ≠ 0 ,

(6) Discuss the continuity and differentiability of dividing throughout by x we get,
f(x) at x = 2
G y Gu Gv Gu Gv
f (x) >x@ if x ∈[0, 4) . [where [*] is a Gx Gx Gx Gx

greatest integer ( floor ) function] Taking the limit as δ x → 0 , we get,

(7) Test the continuity and differentiability of lim § G y · lim § G u G v ·
f(x) = 3x + 2 if x > 2 ¨© G x ¸¹ ©¨ G x G x ¸¹
G xo0 G xo0
= 12 − x2 if x ≤ 2 at x = 2.
lim § G y · lim § G u · lim § G v · ...... (I)
(8) If f(x) = sin x – cos x if x ≤ π / 2 ¨© G x ¸¹ ¨© G x ¹¸ ©¨ G x ¹¸
= 2x − π +1 if x > π / 2 . Test the continuity G xo0 G xo0 G xo0
and differentiability of f at x = π / 2

Since u and v are differentiable function of x

(9) Examine the function § G u · du § G v · dv ….. (II)
©¨ G x ¸¹ dx ©¨ G x ¹¸ dx
 1  ? lim and lim
 x 
f (x) = x2 cos , for x≠0 G xo0 G xo0

= 0 , for x = 0 ∴ lim § G y · du dv [From (I) and (II)]
©¨ G x ¹¸ dx dx
G xo0

for continuity and differentiability at x = 0. i.e. dy du dv
dx dx dx
9.2 RULES OF DIFFERENTIATION

9.2.1. Theorem 1. Derivative of Sum of functions. 9.2.2 Theorem 2. Derivative of Difference of
functions.

If u and v are differentiable functions of x such If u and v are differentiable functions of x such

that y= u+v, then d=y du + dv that y= u − v , then dy du dv
dx dx dx dx dx dx

Proof: Given that, y= u + v where u and v are [Left for students to prove]
differentiable functions of x

188

Corollary : If u, v, w..... are finite number Given that, u and v are differentiable functions of x
of differentiable functions of x such that
y = k1u ± k2v ± k3w ± ..... ? lim § G u · du and lim § G v · dv
¨© G x ¸¹ dx ©¨ G x ¸¹ dx ….. (2)
G xo0 G xo0

then dy k1 du r k2 dv r k3 dw ...... § G y · u dv v du du (0)
dx dx dx dx ¨© G x ¸¹ dx dx dx
? lim

G xo0

9.2.3 Theorem 3. Derivative of Product of [From (1) & (2)]
functions.

If u and v are differentiable functions of x such ? lim § G y · u dv v du
¨© G x ¹¸ dx dx
dy u dv v du G xo0
dx dx dx
that y = u.v , then

Proof : Given that y = uv i.e. dy u dv v du
dx dx dx
Let there be a small increment in the value of x,
Corollary : If u, v and w are differentiable
say δ x , then u changes to (u + δ u) and v changes functions of x such that y = uvw then
to (v + δ v) respectively, correspondingly y
changes to ( y + δ y) dy uv dw uw dv vw du
dx dx dx dx
? y+G y u G u v G v

? G y u Gu v Gv y

? G y uv + uG v vG u G uG v uv 9.2.4 Theorem 4. Derivative of Quotient of
? G y uG v vG u G u.G v functions.

As δ x is small increment in x and δx ≠ 0, If u and v are differentiable functions of x
dividing throughout by δ x We get,
such that

u dy v du u dv
v dx dx dx
Gy uG v vG u G uG v u G v v G v G uG v y = where v ≠ 0 then v2
Gx Gx G x G x Gx

Taking the limit as δ x → 0 , we get, Proof : Given that, y = u , where u and v are
differentiable functions of x v
§ G y · § G v G u G uG v ·
lim ©¨ G x ¹¸ lim ©¨ u G x v G x Gx ¸¹

G xo0 G xo0 Let there be a small increment in the value of x

u lim § G v · v lim § G u · lim § G u · lim G v say δ x then u changes to (u + δ u) and v changes
¨© G x ¸¹ ¨© G x ¸¹ ©¨ G x ¸¹ to (v + δ v) respectively, correspondingly y
G xo0 G xo0 G xo0 G xo0 changes to ( y + δ y)

As δ x → 0 , we get δ v → 0

lim § G y · u lim § G v · v lim § G u · \ y G y u Gu
©¨ G x ¹¸ ¨© G x ¸¹ ¨© G x ¸¹ v Gv
G xo0 G xo0 G xo0

lim § G u · lim G v .... (1) \G y u Gu y
¨© G x ¸¹ v Gv
G xo0 G vo0

189

\G y u Gu u ....... ª  y u º \ SOLVED EXAMPLES
v Gv v ¬« v ¼»
Find the derivatives of the following functions
?G y v(u G u) u(v G v) v.u v.G u u.v u.G v
(v G v).v (v G v).v 3

Ex. 1. 1) y = x2 + logx – cosx

G y v.G u u.G v 2) f(x) = x5 cosecx + x tan x 3) y = ex – 5
(v G v).v ex + 5

As δ x is small increment in x and δ x ≠ 0 , x sin x
dividing throughout by δ x We get, 4) y = x + sin x

Gy v.G u u.G v v. G u u G v Solution :
G x G x
3
G x G x.(v G v).v v2 v.G v
1) Given, y = x2 + logx – cosx

Taking the limit as δ x → 0 , we get, Differentiate w.r.t.x.

§ v. G u u G v · dy = d 3 + logx – cosx)
¨ G x G x ¸ dx dx
§ G y · ¸ ( x2
©¨ G x ¹¸ ¸
lim lim ¨ v2 v.G v
¨
G xo0 G xo0

©¹ = d 3 d (logx) – d (cosx)
dx dx dx
As G x o 0 , we get G v o 0 ( x2 ) +

v. lim § G u · u. lim § G v · = 3 1 1 – (– sin x)
©¨ G x ¸¹ ©¨ G x ¸¹ 2 x
G xo0 G xo0 x2 +

v2 v. lim (G v) ......... (1) dy 3 1
dx 2 x
G vo0 = x + + sin x

Since, u and v are differentiable functions of x 2) Given f(x) = x5 cosecx + x tan x

lim § G u · du and lim § G v · dv Differentiate w.r.t.x.
©¨ G x ¸¹ dx ¨© G x ¸¹ dx ….. (2)
G xo0 G xo0 d
dx
v. du u. dv f ‘(x) = (x5 cosecx + x tan x)
dx dx
lim § G y · ... [ From (1) and (2)] d d
¨© G x ¹¸ dx dx
G xo0 v2 v.(0) = (x5 cosecx) + ( x tan x)

∴ § G y · v du u dv = x5 × d (cosecx) + cosecx × d (x5) +
¨© G x ¸¹ dx dx dx dx
lim
v2
G xo0

v du u dv x × d (tan x) + tan x × d x
dx dx dx dx
i.e. dy v2
dx = x5× (–cosecx × cotx) + cosecx ×(5x4) +
1

x × (sec2x) + tan x × ( )
2x

190

= –x5 cosecx.cotx + 5x4 cosecx + Ex. 2) If f(x) = p tan x + q sin x + r, f(0) = −4 3 ,

x × (sec2x) + 1 tan x §S · §S ·
f ©¨ 3 ¸¹ = −7 3 , f’ ©¨ 3 ¹¸ = 3 then find p, q
2x and r.

3) Given that y = § ex 5·
¨ ex ¸
© 5 ¹ Solution :

Differentiate w.r.t.x. Given that f(x) = p tan x + q sin x + r ...(1)
...(2)
dy d § ex 5 · f’(x) = p sec2 x + q cos x
dx dx
= ¨ ex 5 ¸ f(0) = −4 3
© ¹ put x = 0 in (1)
f(0) = p tan 0 + q sin 0 + r = r ∴r = −4 3
(ex 5) u d (ex 5) (ex 5) u d (ex 5)
dx dx
= (ex 5)2

= (ex 5).(ex ) (ex 5).(ex ) §S ·
(ex 5)2 f ¨© 3 ¹¸ = −7 3 ,

= e2x 5(ex ) e2x 5(ex ) §S · §S ·
(ex 5)2 ∴ from (1) p tan ¨© 3 ¹¸ + q sin ©¨ 3 ¸¹ + r = −7 3

3
p 3 + q − 4 3 = −7 3 ∴ 2p + q = −6 ...(3)

2

= 10 ex §S ·
(ex + 5)2 f’ ¨© 3 ¸¹ = 3

x sin x §S · §S ·
4) y = x + sin x ∴ from (2), psec2 ¨© 3 ¸¹ + qcos ©¨ 3 ¸¹ = 3

Differentiate w.r.t.x., q ...(4)
4p + 2 = 3 ∴ 8p + q = 6
dy d § x sin x ·
dx = dx ©¨ x sin x ¹¸ (4) − (3) gives 6p = 12 ∴ p = 2, put p = 2 in (3),
we get q = −10 ∴ p = 2, q = −10 and r = −4 3
(x sin x) u d (x sin x) (x sin x ) u d (x sin x)
dx dx
= (x sin x)2 9.2.5 Derivatives of Algebraic Functions

(x sin x) u § x d (sin x) sin x d (x) · ( x sin x).(1 cos x) Sr.No. f (x) f '(x)
¨© dx dx ¹¸ 01 c 0
= (x sin x)2 02 xn
nxn-1

(x sin x).(x cos x sin x) (x sin x).(1 cos x) 03 1 − n
= (x sin x)2 04 xn x n +1

= x2 cos x x sin x x sin x cos x sin 2 x x sin x x sin x cos x x
(x sin x)2
1
= x2 cos x + sin2 x 2x
(x + sin x)2

191