Limit and Continuity,_Differentation,_Integration

Lecturer :

Dr. Tang Jing Rui

Prepared by :

Huzainie Hadi Bin Rahman (D20192091464)
Azzah Binti Jamiel @ Jamail (D20192091681)
Izaura Nur Izzaty Binti Ahmad (D20192091885)
Mayang Sari Binti Abdul Wahab (D20192092070)
Mirrah Ayu Farhana Binti Bajuri (D20192092073)

LIMITS AND
CONTINUITY

QUESTION (C1)
SOLUTION

QUESTION Infinite and/or Nonexistent Limits (C2)

SOLUTION

SOLUTION

QUESTION A Nonexistent Limit (C2)

SOLUTION

QUESTION (C3)
SOLUTION

SOLUTION

QUESTION SOLUTION (C2)

Find the following limits. If a limit does not exist, say (C2)
so, and use the symbols ∞ if appropriate.

QUESTION SOLUTION

Find the following limits. If a limit does not exist, say (C3)
so, and use the symbols ∞ if appropriate.

QUESTION SOLUTION

Find lim f ( x ) , where x→1 2 − 1 − 1 + 1
= − 1 =
f ( x ) = ൜− 2−1 = + 1 − 1

−1 ∴≠ 1

x≠1 lim = lim + 1 = 1 + 1 = 2

x=1 x→1 x→1

QUESTION (C3)
SOLUTION

SOLUTION

Find the following limits. If a limit does not exist, say (C1)
so, and use the symbols ∞ if appropriate.

QUESTION SOLUTION

lim (x² + 3x – 7) This is same function as limit at
x→3 x=3 will be the same as the value
of the function at x=3

Substitute in x = 3
= 3² + 3*3 – 7
= 11

Consider the function

(C4)

QUESTION SOLUTION

[ 1 if 0 ≤ x < 1 a) lim f(x)
x → 1+
f(x) = x² if x ≥ 1
= lim x²
a) Find f(x) x → 1+
lim f(x)
x → 1+ =1
lim lim f(x)
x → 1¯ x → 1¯

b) What does this tell you about = lim 1
lim f(x)? x → 1¯
x→1
=1
c) Is f(x) continuous at x=1 ? b) The limit exists and is equal to 1
c) Yes, since f(1) = 1² = 1 so
f(1) = lim f(x)

x→1
Thus f(x) is continuous at x = 1

Consider the function f(x) = | x | - x

(C2)

QUESTION SOLUTION

a) (i) What is f(x) when x > 0 ? a) (i) Since x > 0, |x| = x
(ii) What is f(x) when x < 0 ? so f(x) = x – x = 0

(ii) Since x < 0, |x| = -x
f(x) = -x – x = -2x

b) (i) Find lim f(x) b) (i) Find lim f(x) since x>0,
x → 0+ f(x) x → 0+ f(x) = 0

(ii) Find lim = lim 0
x → 0¯ x → 0+

=0

Consider the function f(x) = | x | - x

(C1)

QUESTION SOLUTION

Consider the function f(x) = | x | - x

QUESTION SOLUTION (C2)

Consider the function f(x) = | x | - x

QUESTION SOLUTION (C2)

DIFFERENTATION

QUESTION SOLUTION (C2)

QUESTION SOLUTION (C2)

QUESTION SOLUTION (C1)

QUESTION SOLUTION (C1)

First Derivative of a Composite Function (C3)

QUESTION

SOLUTION

QUESTION (C3)
SOLUTION

QUESTION (C4)
SOLUTION

QUESTION (C4)

SOLUTION

QUESTION SOLUTION (C1)

ⁿ = ⁿ¯1


Find the derivative of f(x) ( 2 x⁹) = 2 9
3 3
= 2 ⁹ 2
3 = 3 (9 9¯1)

= 2 9 8
3
= 6x⁸

QUESTION SOLUTION (C2)

Calculate the derivative of 2 3 − 4 2 + − 33
f(x) = 2x³ - 4x² + x – 33

2 3 4 2
= − +


− 33

3 2
= 2 − 4 +

= 2 3 2 − 4 2 1 + 1

= 6x2 − 8 + 1

QUESTION SOLUTION (C1)

Differentiate f(x) = 5 5
³ 3
= 5 ¯³
= 5 −3 (¯³¯¹⁾
= −15 ¯4
−15
= ⁴

QUESTION (C1)
SOLUTION

QUESTION SOLUTION (C2)

QUESTION SOLUTION (C1)

QUESTION SOLUTION (C3)

INTEGRATION

QUESTION SOLUTION (C2)

SOLUTION

QUESTION SOLUTION (C2)

QUESTION (C3)

SOLUTION

QUESTION SOLUTION (C4)

QUESTION SOLUTION (C4)

QUESTION (C2)
SOLUTION

QUESTION (C3)
SOLUTION

QUESTION (C3)
SOLUTION

QUESTION SOLUTION (C3)

Let u = x²

Then = 2

1
Use a suitable So dx = 2

substitution to find the Making the substitution :

following integrals ʃ x.cos(u) 1
2
ʃxcos(x²)dx 1
= 2 ʃ cos

= 1 sin +
2
1 sin
= 2 +
2

QUESTION SOLUTION (C3)

Use the formula Let = cos

ʃ u = . − ʃ
Then u = x

= = 1 = sin
To find the following
integrals
i) ʃ xcos(x)dx
Substituting into the integration by
parts formula:
ʃ xcos (x)dx
= xsin(x) - ʃsin(x).1dx
= xsin(x) + cos(x) + C