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SCHEME OF WORK Week Topics 1 1. 0 FUNCTIONS, LIMITS AND CONTINUITY 1.1 Functions ▪ Properties of functions ▪ Domain and range ▪ Operations on functions ▪ Graphs (Family) of functions Lecture Discussion Tutorial/lab 2 1.2 Limits ▪ Introduction (An intuitive approach) ▪ One-sided limit and two-sided limits ▪ Infinite limits and limits at infinity ▪ Computing limits Lecture Discussion Tutorial/lab 3 ▪ Computing limits: End behavior 1.3 Continuity ▪ Continuity at a point ▪ Limits and continuity of trigonometric functions Lecture Discussion Tutorial/lab 4 1.0 DIFFERENTIATION 2.1 An Introduction to the Derivative: Tangents 2.2 The Definition of Derivative ▪ Differentiation from the first principle ▪ Differentiability and continuity 2.3 Techniques of Differentiation 2.4 Derivatives of Trigonometric, Exponential and Logarithmic Functions Lecture Discussion Tutorial/lab 5 2.5 The Chain Rule 2.6 Implicit Differentiation 2.7 Linear Approximations and Differentials Lecture Discussion Tutorial/lab 6 3.0 APPLICATIONS OF DIFFERENTIATION 3.1 Related Rates Lecture Discussion Tutorial/lab QUIZ
Week Topics 7 3.2 Analysis of Functions I: ▪ Intervals of increasing and decreasing functions ▪ Concavity and inflection points 3.3 Analysis of Functions II: ▪ Relative maxima and minima Lecture Discussion Tutorial/lab 8 ▪ Critical numbers ▪ First and Second Derivative Test ▪ Graphs of polynomial functions ▪ Graphs of rational functions (linear /linear) Lecture Discussion Tutorial/lab 9 3.4 Maximum and Minimum Values ▪ Absolute maxima and minima 3.5 Applied Maximum and Minimum Problems 3.6 Rolle’s Theorem; Mean Value Theorem Lecture Discussion Tutorial/lab 10 4.0 INTEGRATION 4.1 Antiderivatives 4.2 The indefinite integral of algebraic functions and trigonometric functions Exponential and Logarithmic Functions 4.3 Integration by Substitution Lecture Discussion Tutorial/lab 11 4.4 Sigma Notation 4.5 Area as a limit and the Definite Integrals ▪ Properties of Definite Integrals Lecture Discussion Tutorial/lab TEST 12 4.6 The Fundamental Theorem of Calculus ● The Fundamental Theorem of Calculus, Part 1 ● The Fundamental Theorem of Calculus, Part 2 ● Evaluating Definite Integrals by Substitution 4.7 The Mean Value Theorem for Integrals Lecture Discussion Tutorial/lab
Week Topics 13 1.0 APPLICATIONS OF INTEGRATION 1.1 Area Between Two Curves 1.2 Volume: Solids of Revolution ● Volume by Disks Method Lecture Discussion Tutorial/lab 14 5.2 Volume: Solids of Revolution ● Volume by Washers Method 5.3 Volume by Cylindrical Shells Method Lecture Discussion Tutorial/lab
Chapter 1 : Functions, Limits & Continuity 1 CHAPTER 1 : FUNCTIONS, LIMITS AND CONTINUITY The concept of the Limits and Continuity is one of the most crucial things to understand in order to prepare for calculus. Who invented calculus? Gottfried Leibnitz is a famous German philosopher and mathematician, and he was a contemporary of Isaac Newton. These two gentlemen are the founding fathers of Calculus, and they did most of their work in 1600s. We still use the Leibniz notation of dx dy for most purposes. LIMITS A limit is a number that a function approaches as the independent variable of the function approaches a given value. For example, given the function f (x) = 3x , you could say, “The limit of f (x) as x approaches 2 is 6 .” Symbolically, this is written lim ( ) 6 2 = → f x x . Theorem of Limits Properties Examples a. c c x a = → lim lim 2 = 2 x→a b. n n x a x = a → lim 1. lim 2 2 = → x x 2. lim 2 8 3 3 2 = = → x x 3. = + →+ x x lim 4. = − →− x x lim c. lim [f(x) g(x)] lim f(x) lim g(x) x→a x→a x→a = 12 3 3 lim [ ] lim lim 2 3 2 3 2 3 = = + + = + → → → x x x x x x x d. lim [f(x) g(x)] lim f(x) lim g(x) x→a x→a x→a • = • 4 1 (3 1) lim (3 ) lim lim(3 ) 2 1 2 1 2 1 = = + + = • + → → → x x x x x x x e. , lim ( ) 0 lim ( ) lim ( ) ( ) ( ) lim = → → → → g x g x f x g x f x x a x a x a x a 7 4 lim 3 lim 2 3 2 lim 2 2 2 2 2 = + = + → → → x x x x x x x f. lim [cf(x)] c lim f(x) x→a x→a = 14 2(7) lim 2( 4) 2 lim ( 4) 3 3 = = + = + → → x x x x g. lim ( ) = lim ( ), lim ( ) 0 → → → f x f x f x x a n x a n x a n = positive integer. 3 81 lim 2 7 51 lim (2 7 51) 4 4 3 2 4 3 2 = = + + = + + → → x x x x x x
MAT183/ MAT421 : Calculus I 2 Methods of Evaluating Limits i) Determinate Forms of Limits (Limits by Direct Substitution) To find f(x) x→a lim , we substitute x = a in the function. If the value x →a comes out to be definite value, it is limit. That is f(x) f(a) x a = → lim provided is exists. ii) Indeterminate Forms of Limits (Cannot use Direct Substitution) Be careful when a quotient is involved. If direct substitution of x = a while evaluating ( ) g(x) f x x→a lim gives 0 0 or , then we cannot use direct substitution anymore. ( ) ( ) 0 0 lim = → g x f x x a In limits, let say after we use direct substitution, the final answer gives 0 0 ; then there are 3 cases. Case 1 : f(x) or g(x) are quadratic function – factorize, simplify, subs the value of x. Case 2 : f(x) or g(x) are surd ( ) function – multiplying by their conjugate, expand and simplify using (a + b)(a − b) = a 2 − b 2 , simplify, subs the value of x. Case 3 : f(x) or g(x) are trigonometric functions – use squeezing theorem*, subs the value of x. *squeezing theorem states that : i) 1 sin lim 0 = → x x x Or 1 2 sin2 lim 0 = → x x x , 1 3 sin3 lim 0 = → x x x , 1 4 sin4 lim 0 = → x x x , etc… ii) 0 1 cos lim 0 = − → x x x note : 1. when we use squeezing theorem, if the denominator not equal to the angle of sine, we need to adjust the denominator until it is exactly equal to the angle of sine 2. never adjust the angle of sine in order to be equal to the denominator. Note : when we substitute the value of x into variable x, no more limits expressions (to write)
Chapter 1 : Functions, Limits & Continuity 3 ( ) ( ) = →+ g x f x x lim (positive infinite limits) Case 1 : f(x) or g(x) are polynomial – dividing each expression (numerator and denominator) by the highest power of x from denominator, simplify, subs the value of x by and use a = 0 (where a = constant) which means : any numbers divide by will approach to zero. Case 2 : f(x) or g(x) are surd ( ) function – dividing each expression (numerator and denominator) by the highest power of x from denominator, write x = 2 x or x 2 = 4 x (etc..), simplify using b a b a = , subs the value of x by and use a = 0 (where a = constant) which means : any numbers divide by will approach to zero. ( ) ( ) = →− g x f x x lim (negative infinite limits) Case 1 : f(x) or g(x) are polynomial – dividing each expression (numerator and denominator) by the negative highest power of x from denominator, simplify, subs the value of x by and use a = 0 (where a = constant) which means : any numbers divide by will approach to zero. Case 2 : f(x) or g(x) are surd ( ) function – dividing each expression (numerator and denominator) by the negative highest power of x from denominator, write x = 2 x or x2 = 4 x (etc..), simplify using b a b a = , subs the value of x by and use a = 0 (where a = constant) which means : any numbers divide by will approach to zero.
MAT183/ MAT421 : Calculus I 4 Examples for ( ) ( ) 0 0 = → g x f x lim x a , Quadratic or polynomial Functions (factorize) a) Evaluate 12 16 4 lim 2 4 − − − → x x x x . How about 12 16 4 lim 2 0 − − − → x x x x ? Solution : Using direct substitution, →4 16 − 4 2 − − 12 = 0 0 ( ) ( ) Thus, lim→4 16 − 4 2 − − 12 = →4 4(4 − ) ( + 3)( − 4) = →4 −4( − 4) ( + 3)( − 4) = →4 −4 ( + 3) = −4 (4 + 3) = − 4 7 # Using direct substitution, →0 16 − 4 2 − − 12 = 16 −12 = − 4 3 () or, →0 16 − 4 2 − − 12 = →0 4(4 − ) ( + 3)( − 4) = →0 −4( − 4) ( + 3)( − 4) = →0 −4 ( + 3) = −4 (0 + 3) = − 4 3 #
Chapter 1 : Functions, Limits & Continuity 5 b) Evaluate 5 3 10 2 5 + + − →− x x x lim x (ans : −7) Solution : c) Evaluate 4 12 4 2 2 2 + − − → x x x lim x (ans : 2 1 ) Solution : Using direct substitution, →−5 2 + 3 − 10 + 5 = 0 0 ( ) ( ) Thus, →−5 2 + 3 − 10 + 5 = →−5 ( − 2)( + 5) + 5 = →−5 ( − 2) = (−5 − 2) = −7 # Using direct substitution, →2 2 − 4 2 + 4 − 12 = 0 0 ( ) ( ) Thus, →2 2 − 4 2 + 4 − 12 = →2 ( + 2)( − 2) ( − 2)( + 6) = →2 ( + 2) ( + 6) = (2 + 2) (2 + 6) = 4 8 = 1 2 #
MAT183/ MAT421 : Calculus I 6 d) Evaluate 4 2 3 2 0 3 2 2 3 lim t t t t t − + → (ans : − 2 3 ) Solution : e) Evaluate x x x x e e e − − → 1 lim 2 0 (ans : −1) Solution : Note : After direct substitution, the answer gives 0 0 . There is no surd and no trigonometric functions → need to factorize. lim→0 2− 1 − = lim→0 ( ) 2 − 1 − = lim→0 2 − 1 − = lim→0 ( − 1) 1 − = lim→0 ( − 1) −( − 1) = −lim→0 = −lim→0 = − 0 = − 1 # ( ) = = Where = Factorize −( − 1) = − + 1 = ( − 1) Replace by Using direct substitution, →0 2 3 + 3 2 3 4 − 2 2 = 0 0 ( ) ( ) Thus, →0 2 3 + 3 2 3 4 − 2 2 = →0 2 (2 + 3) 2(3 2 − 2) = →0 (2 + 3) (3 2 − 2) = (0 + 3) (0 − 2) = − 3 2 #
Chapter 1 : Functions, Limits & Continuity 7 Surd (multiplying with their conjugate) Use (a + b)(a − b) = a 2 − b 2 What is Conjugate?? And why we multiplying with their conjugate?? Examples a) Evaluate 2 4 lim 4 − − → x x x Solution : expression conjugate multiplying with their conjugate x − 2 x + 2 ( 2)( 2) ( ) (2) 4 2 2 x − x + = x − = x − x + 3 x − 3 ( 3)( 3) ( ) (3) 9 2 2 x + x − = x − = x − x + 3 − 2 x + 3 + 2 ( )( ) ( ) ( ) 1 3 4 3 2 3 2 3 2 2 2 = − = + − + − + + = + − x x x x x Multiplying with their conjugate Multiplying with their conjugate conjugate conjugate Free from surd symbol Free from surd symbol Multiplying with their conjugate conjugate Free from surd symbol →4 − 4 √ − 2 = →4 − 4 √ − 2 ∙ √ + 2 √ + 2 = →4 ( − 4)(√ + 2) (√ − 2)(√ + 2) = →4 ( − 4)(√ + 2) (√) 2 − (2) 2 = →4 ( − 4)(√ + 2) − 4 = →4 (√ + 2) = √4 + 2 = 2 + 2 = 4 #
MAT183/ MAT421 : Calculus I 8 b) Evaluate x x x − + − → 1 3 2 lim 1 Solution : →1 √ + 3 − 2 1 − = →1 √ + 3 − 2 1 − • √ + 3 + 2 √ + 3 + 2 = →1 (√ + 3 − 2)(√ + 3 + 2) (1 − )(√ + 3 + 2) = →1 (√ + 3) 2 − (2) 2 (1 − )(√ + 3 + 2) = →1 + 3 − 4 (1 − )(√ + 3 + 2) = →1 − 1 (1 − )(√ + 3 + 2) = →1 −(1 − ) (1 − )(√ + 3 + 2) = →1 −1 √ + 3 + 2 = −1 √1 + 3 + 2 = −1 √4 + 2 = −1 2 + 2 = − 1 4 #
Chapter 1 : Functions, Limits & Continuity 9 c) Evaluate 1 8 3 lim 2 1 + + − →− x x x Solution : →−1 √ 2 + 8 − 3 + 1 = →−1 √ 2 + 8 − 3 + 1 • √ 2 + 8 + 3 √ 2 + 8 + 3 = →−1 (√ 2 + 8 − 3)(√ 2 + 8 + 3) ( + 1)(√ 2 + 8 + 3) = →−1 (√ 2 + 8) 2 − (3) 2 ( + 1)(√ 2 + 8 + 3) = →−1 2 + 8 − 9 ( + 1)(√ 2 + 8 + 3) = →−1 2 − 1 ( + 1)(√ 2 + 8 + 3) = →−1 ( + 1)( − 1) ( + 1)(√ 2 + 8 + 3) = →−1 − 1 √ 2 + 8 + 3 = −1 − 1 √1 + 8 + 3 = −2 √9 + 3 = −2 3 + 3 = − 2 6 = − 1 3 #
MAT183/ MAT421 : Calculus I 10 d) Evaluate x x x 2 1 4 1 lim 0 − + → Solution : →0 1 √ + 4 − 1 2 = →0 1 • ( 1 √ + 4 − 1 2 ) = →0 1 • ( 2 − √ + 4 2√ + 4 ) = →0 2 − √ + 4 2√ + 4 = →0 2 − √ + 4 2√ + 4 • 2 + √ + 4 2 + √ + 4 = →0 (2 − √ + 4)(2 + √ + 4) 2√ + 4(2 + √ + 4) = →0 (2) 2 − (√ + 4) 2 2√ + 4(2 + √ + 4) = →0 4 − ( + 4) 2√ + 4(2 + √ + 4) = →0 4 − − 4 2√ + 4(2 + √ + 4) = →0 − 2√ + 4(2 + √ + 4) = →0 −1 2√ + 4(2 + √ + 4) = −1 2√4(2 + √4) = −1 2 • 2(2 + 2) = −1 2 • 2 • 4 = − 1 16 #
Chapter 1 : Functions, Limits & Continuity 11 Trigonometric Function : use Squeezing theorem 1 0 = → x sin x lim x and 0 1 cos lim 0 = − → x x x In general, (from 1 0 = → x sin x lim x ) if the angle of sine is exactly equal to the denominator, we still can use the squeezing theorem as follows, 1 0 = → x sin x lim x 1 2 2 0 = → x sin x lim x 1 3 3 0 = → x sin x lim x 1 4 4 0 = → x sin x lim x 1 5 5 0 = → x sin x lim x etc.. How about if the angle of sine is not equal to the denominator such as x sin x lim x 2 →0 ? The solution should be like this, x sin x lim x 2 →0 = x x x 2 2sin2 lim →0 = 2 x x x 2 sin2 lim →0 = 2(1) = 2 # (this is correct and acceptable) Caution : if the angle of sine is not equal to the denominator • never adjust the angle of sine • adjust denominator until the denominator is equal to the angle of sine Common mistake x sin x lim x 2 →0 = x x x 2sin lim →0 = 2 x x x sin lim →0 = 2(1) = 2 # (this is wrong and not acceptable)
MAT183/ MAT421 : Calculus I 12 Example 1 : Find x sin x lim x 3 →0 Solution : Example 2 : Find sin x sin x lim x 2 3 →0 Solution : →0 3 = →0 3 3 3 = 3 →0 3 3 = 3(1) = 3 # →0 3 2 = →0 3 2 • 1 ⁄ 1 ⁄ = →0 ( 3 ) ( 2 ) = →0 ( 3 ) →0 ( 2 ) = 3→0 ( 3 3 ) 2→0 ( 2 2 ) = 3(1) 2(1) = 3 2 #
Chapter 1 : Functions, Limits & Continuity 13 Example 3 : Find sin x tan x lim x 3 7 →0 Solution : →0 7 3 = →0 ( 7 7 ) 3 = →0 [( 7 7 ) ÷ 3] = →0 [ 7 7 • 1 3 ] = →0 [ 7 3 • 1 7 ] = →0 7 3 • →0 1 7 = →0 7 3 • ( 1 ) ( 1 ) • →0 1 7 = →0 ( 7 ) ( 3 ) • →0 1 7 = →0 ( 7 ) →0 ( 3 ) • 1 7(0) = 7 →0 ( 7 7 ) 3 →0 ( 3 3 ) • 1 (0) = 7(1) 3(1) • 1 1 = 7 3 #
MAT183/ MAT421 : Calculus I 14 Example 4 : Find cot x cot x lim x 5 →0 Solution : →0 5 = →0 ( 5 5 ) ( ) = →0 ( 5 5 ÷ ) = →0 ( 5 5 • ) = →0 ( 5 • 5 ) = →0 ( 5 ) • →0 ( 5 ) = →0 ( 5 ) • →0 ( 5 ) • ( 1 ) ( 1 ) = 5(0) (0) • →0 ( ) ( 5 ) = 1 1 • →0 ( ) →0 ( 5 ) = →0 ( ) 5 →0 ( 5 5 ) = 1 5(1) = 1 5 #
Chapter 1 : Functions, Limits & Continuity 15 Example 5 : Find lim sin x x 2 →0 Solution : Example 6 : Find lim cos x x 3 →0 Solution : Example 7 : Find x x x x 4sin7 lim 3 0 + → Solution : →0 3 + 4 7 = →0 ( 3 + 4 7 ) = →0 ( 2 + 4 7 ) = →0 ( 2 ) + →0 ( 4 7 ) = →0 2 + 4→0 ( 7 ) = 0 2 + 4(7)→0 ( 7 7 ) = 0 + (28)(1) = 28 # →0 2 = 2(0) = (0) = 0 # →0 3 = 3(0) = (0) = 1 #
MAT183/ MAT421 : Calculus I 16 Example 8 : Find x x x 2 sin lim →0 Solution : Separate using theorem of limits lim→0 2√ = lim→0 2√ ∙ √ √ = lim→0 √ 2(√) 2 = lim→0 √ 2 = lim→0 ( √ 2 ∙ ) = lim→0 ( √ 2 ) ∙ lim→0 ( ) = ( √0 2 ) ∙ (1) = (0) ∙ (1) = 0 # Use direct subs. (no more limits) Use squeezing theorem
Chapter 1 : Functions, Limits & Continuity 17 Example 9 : Find →0 2−2 3 2 Solution : Example 10 : Find →0 ( 2 ) cos Solution : →0 2 − 2 3 2 = →0 2 3 2 − →0 2 3 2 = →0 1 3 − 1 3 →0 ( ∙ ) = 1 3 − 1 3 (→0 ∙ →0 ) = 1 3 − 1 3 (1 ∙ 1) = 1 3 − 1 3 = 0 # →0 ( 2 ) cos = →0 ( ( 2 ) ∙ 1 ) = →0 ( 2 ) ∙ →0 1 cos = 1 2 →0 ( 2 ) 1 2 ∙ ( 1 0 ) = 1 2 →0 ( 2 ) ( 2 ) ∙ ( 1 1 ) = 1 2 (1) ∙ ( 1 1 ) = 1 2 # Separate the limits Use squeezing theorem Use direct substitution (no more limit)
MAT183/ MAT421 : Calculus I 18 Limits to Infinity Infinity is a very special idea. We know we can't reach it, but we can still try to work out the value of functions that have infinity in them. One Divided by Infinity Let's start with an interesting example. Question: What is the value of 1 ∞ ? Answer: We Can Approach It! So instead of trying to work it out for infinity (because we can't get a sensible answer), let's try larger and larger values of x: x x 1 1 1.00000 2 0.50000 4 0.25000 10 0.10000 100 0.01000 1,000 0.00100 10,000 0.00010 Now we can see that as x gets larger, 1 tends towards 0 We are now faced with an interesting situation: • We can't say what happens when x gets to infinity • But we can see that 1 is going towards 0 • We want to give the answer "0 “but can’t, so instead mathematicians say exactly what is going on by using the special word” limits The limit of 1 as x approaches Infinity is 0
Chapter 1 : Functions, Limits & Continuity 19 And write it like this: In other words: As x approaches infinity, then 1 approaches 0 When you see "limit", think "approaching" It is a mathematical way of saying "we are not talking about when x=∞, but we know as x gets bigger, the answer gets closer and closer to 0". Examples : i) 0 1 1 lim = x→ x ii) 0 1 1 1 lim 2 2 = = x→ x iii) 0 1 1 1 lim 3 3 = = x→ x iv) 0 3 3 lim = x→ x v) 0 4 4 4 lim 2 2 = = x→ x vi) 0 2 2 2 lim 5 5 = = x→ x Theorem of Infinite Limits : 0 1 → n x x lim , where n = 1, 2, 3, …
MAT183/ MAT421 : Calculus I 20 Examples for ( ) ( ) = → g x f x lim x Positive and Negative Infinite Limits involving polynomial functions – we need to simplify the function by dividing the numerator and denominator by the highest power of x that occurs in the denominator. →+∞ 4 5 − + 1 3 4 − 2 2 = →+∞ 4 5 3 4 = 4 3 →+∞ 5 4 = 4 3 →+∞ = 4 3 (∞) = ∞ # 1.a) i. Take the term containing the highest power of x for numerator and denominator 1.a) i. 1.a) i. →∞ 4 5 − + 1 3 4 − 2 2 = →∞ ( 4 5 − + 1 4 ) ( 3 4 − 2 2 4 ) = →∞ 4 5 4 − 4 + 1 4 34 4 − 22 4 = →∞ 4 − 1 3 + 1 4 3 − 2 2 = 4(∞) − 1 ∞3 + 1 ∞4 3 − 2 ∞2 = 4(∞) − 0 + 0 3 − 0 = 4(∞) 3 = −∞ # Dividing the numerator and denominator by the highest power of x from denominator Factorize the highest power of x for numerator and denominator →∞ 4 5 − + 1 3 4 − 2 2 = →∞ 5 (4 − 1 4 + 1 5) 4(3 − 2 2) = →∞ (4 − 1 4 + 1 5) (3 − 2 2) = ∞(4 − 1 ∞4 + 1 ∞5) (3 − 2 ∞2) = ∞(4 − 0 + 0) (3 − 0) = ∞(4) 3 = ∞ #
Chapter 1 : Functions, Limits & Continuity 21 1.a) ii. 1.a) ii. 1.a) ii. →−∞ 4 5 − + 1 3 4 − 2 2 = →−∞ ( 4 5 − + 1 − 4 ) ( 3 4 − 2 2 − 4 ) = →−∞ 4 5 − 4 − − 4 + 1 − 4 34 −4− 22 −4 = →−∞ −4 + 1 3 − 1 4 −3 + 2 2 = −4(−∞) + 1 (−∞) 3 − 1 (−∞) 4 −3 + 2 (−∞) 2 = −4(−∞) − 0 − 0 −3 + 0 = −4(−∞) −3 = −∞ # Dividing the numerator and denominator by the highest power of x from denominator Factorize the highest power of x for numerator and denominator →−∞ 4 5 − + 1 3 4 − 2 2 = →−∞ 5 (4 − 1 4 + 1 5) 4(3 − 2 2) = →−∞ (4 − 1 4 + 1 5) (3 − 2 2) = −∞ (4 − 1 (−∞) 4 + 1 (−∞) 5) (3 − 2 (−∞) 2) = −∞(4 − 1 ∞4 − 1 ∞5) (3 − 2 ∞2) = −∞(4 − 0 − 0) (3 − 0) = −∞(4) 3 = −∞ # →−∞ 4 5 − + 1 3 4 − 2 2 = →−∞ 4 5 3 4 = 4 3 →−∞ 5 4 = 4 3 →−∞ = 4 3 (−∞) = −∞ # Take the term containing the highest power of x for numerator and denominator
MAT183/ MAT421 : Calculus I 22 • Never use Quick Method to solve limits x approaches to any real number • Quick method only can use to solve Limits at Infinity 1.b) 1.b) →−∞ ( 3 + 1 ) ( 4 2 + 1 2 2 + ) 3 = Dividing the numerator and denominator by the highest power of x from denominator →−∞ ( 3 + 1 ) ( 4 2 + 1 2 2 + ) 3 = →−∞ ( 3 ) ( 4 2 2 2 ) 3 = →−∞ ( 1 3 ) (2) 3 = →−∞ ( 8 3 ) = 8 3 # Take the term containing the highest power of x for numerator and denominator
Chapter 1 : Functions, Limits & Continuity 23 More Examples Example 1 Let h(x) = 2 1 17 2 2 − + + x x x , find h(x) h(x) x→ x→− lim + lim 2 . Solution : →2 ℎ() = →2 17 + 2 2 − 2 + 1 = 17 + 2(2) (2) 2 − 2(2) + 1 = 17 + 4 4 − 4 + 1 = 21 →−∞ ℎ() = →−∞ 17 + 2 2 − 2 + 1 = →−∞ ( 17 + 2) 2(1 − 2 + 1 2) = →−∞ 17 + 2 (1 − 2 + 1 2) = 17 −∞ + 2 −∞ (1 − 2 −∞ + 1 (−∞) 2) = −0 + 2 −∞(1 + 0 + 0) = 2 −∞ = 0 Hence, →2 ℎ() + →−∞ ℎ() = 21 + 0 = 21 #
MAT183/ MAT421 : Calculus I 24 Example 2 Let h(x) = 3 2 3 3 9 x x x x − − , find h(x) h(x) x→ x→− lim − lim 3 . Solution : →3 ℎ() = →3 3 − 9 3 − 3 2 = →3 ( 2 − 9) 2( − 3) = →3 ( 2 − 9) ( − 3) = →3 ( + 3)( − 3) ( − 3) = →3 ( + 3) = (3 + 3) 3 = 6 3 = 2 →−∞ ℎ() = →−∞ 3 − 9 3 − 3 2 = →−∞ 3 (1 − 9 2) 3(1 − 3 ) = →−∞ 1 − 9 2 1 − 3 = 1 − 9 (−∞) 2 1 − 3 −∞ = 1 − 0 1 + 0 = 1 Hence, →3 ℎ() − →−∞ ℎ() = 2 − 1 = 1 #
Chapter 1 : Functions, Limits & Continuity 25 Example 3 Let g(x) = x x x x − + − 2 2 2 , find g(x) g(x) x→ x→+ lim + lim 1 . Solution : →1 () = →1 2 + − 2 2 − = →1 ( − 1)( + 2) ( − 1) = →1 ( + 2) = (1 + 2) 1 = 3 →+∞ () = →+∞ 2 + − 2 2 − = →+∞ 2 (1 + 1 − 2 2) 2(1 − 1 ) = →+∞ (1 + 1 − 2 2) (1 − 1 ) = 1 + 1 ∞ − 2 ∞2 1 − 1 ∞ = 1 + 0 − 0 1 − 0 = 1 Hence, →1 () + →+∞ () = 3 + 1 = 4 #
MAT183/ MAT421 : Calculus I 26 Positive and Negative Infinite Limits involving Radicals 2.a) 6 3 6 11 25 5 lim x x x x − + → Solution : →∞ √ 25 6 + 5 6 − 11 3 = √→∞ 25 6 + 5 6 − 11 3 = √→∞ ( 25 6 + 5 6 ) ( 6 − 11 3 6 ) = √→∞ 25 6 ⁄ 6 + 5 ⁄ 6 6 ⁄ 6 − 11 3 ⁄ 6 = √→∞ 25 + 5 ⁄ 6 1 − 11 ⁄ 3 = √ 25 + 5⁄∞6 1 − 11⁄∞3 = √ 25 + 5⁄∞ 1 − 11⁄∞ = √ 25 + 0 1 − 0 = √25 = 5 # →∞ √ 25 6 + 5 6 − 11 3 = √→∞ 25 6 6 = √→∞ 25 = √25 = 5 # →∞ √ 25 6 + 5 6 − 11 3 = √→∞ 25 6 + 5 6 − 11 3 = √→∞ 6 (25 + 5 6 ) 6 (1 − 11 3 ) = √→∞ (25 + 5 6 ) (1 − 11 3 ) = √ 25 1 = √25 = 5 #
Chapter 1 : Functions, Limits & Continuity 27 2. ) () →+∞ √2 2 − 3 + 2 Solution : →+∞ √2 2 − 3 + 2 = →+∞ √ 2(2 − 3 2) (1 + 2 ) = →+∞ √ 2√2 − 3 2 (1 + 2 ) = →+∞ ||√2 − 3 2 (1 + 2 ) = →+∞ √2 − 3 2 (1 + 2 ) = →+∞ √2 − 3 2 (1 + 2 ) = √2 − 3 (∞) 2 (1 + 2 ∞ ) = √2 − 0 (1 + 0) = √2 1 = √2 # 2. ) () →−∞ √2 2 − 3 + 2 Solution : →−∞ √2 2 − 3 + 2 = →−∞ √ 2(2 − 3 2) (1 + 2 ) = →−∞ √ 2√2 − 3 2 (1 + 2 ) = →−∞ ||√2 − 3 2 (1 + 2 ) = →−∞ −√2 − 3 2 (1 + 2 ) = →−∞ −√2 − 3 2 (1 + 2 ) = −√2 − 3 (−∞) 2 (1 + 2 −∞ ) = −√2 − 3 ∞2 (1 − 2 ∞ ) = −√2 − 0 (1 − 0) = −√2 1 = −√2 # ඥ = || = ≥ − <
MAT183/ MAT421 : Calculus I 28 2. ) ) →−∞ √3 4 + 2 4 2 − 3 Solution : →−∞ √3 4 + 2 4 2 − 3 = →−∞ √ 4(3 + 2 4) 2(4 − 3 2) = →−∞ √ 4√3 + 2 4 2(4 − 3 2) = →−∞ | 2 |√3 + 2 4 2(4 − 3 2) = →−∞ 2√3 + 2 4 2(4 − 3 2) = →−∞ √3 + 2 4 (4 − 3 2) = √3 + 2 (−∞) 4 (4 − 3 (−∞) 2) = √3 + 2 ∞4 (4 − 3 ∞2) = √3 + 0 (4 − 0) = √3 4 # 2. ) ) →+∞ √3 4 + 2 4 2 − 3 Solution : →+∞ √3 4 + 2 4 2 − 3 = →+∞ √ 4(3 + 2 4) 2(4 − 3 2) = →+∞ √ 4√3 + 2 4 2(4 − 3 2) = →+∞ | 2 |√3 + 2 4 2(4 − 3 2) = →+∞ 2√3 + 2 4 2(4 − 3 2) = →+∞ √3 + 2 4 (4 − 3 2) = √3 + 2 (∞) 4 (4 − 3 (∞) 2) = √3 + 0 (4 − 0) = √3 4 # √ = ห ห = − ∞ < < +∞
Chapter 1 : Functions, Limits & Continuity 29 More Examples Example 1 Find →−∞ √3 2− 4−5 . Solution : →−∞ √3 2 − 4 − 5 = →−∞ √ 2(3 − 1 ) (4 − 5 ) = →−∞ √ 2√3 − 1 (4 − 5 ) = →−∞ ||√3 − 1 (4 − 5 ) = →−∞ −√3 − 1 (4 − 5 ) = →−∞ −√3 − 1 (4 − 5 ) = −√3 − 1 −∞ (4 − 5 −∞ ) = −√3 + 1 ∞ (4 + 5 ∞ ) = −√3 + 0 (4 + 0) = − √3 4 # ඥ = || = ≥ − <
MAT183/ MAT421 : Calculus I 30 Example 2 Find x x x x 6 4 3 lim 8 8 − + → . Ans : 2 Solution : Two Sided limits If f(x) L x a = → − lim and f(x) L x a = → + lim then f(x) L x a = → lim (exist) Or f(x) L x a = → lim exist, if and only if f(x) x a → − lim = f(x) x a → + lim f(x) x a → − lim f(x) x a → + lim , then f(x) x→a lim does not exist Lim x to a from the left Lim x to a from the right →∞ √ 4 8 + 3 8 − 6 = →∞ √ 8(4 + 3 8) 8(1 − 6 7) = →∞ √ 4 + 3 8 1 − 6 7 = √ 4 + 3 ∞8 1 − 6 ∞7 = √ 4 + 0 1 − 0 = √4 = 2 #
Chapter 1 : Functions, Limits & Continuity 31 Example 1 : f(x) single function (limit always exist) Given f(x) = 3 9 2 − − x x . Determine whether →3 () exist or does not exist. Solution : Example 2 : g(x) piece-wise function (limit always exist) Given () = { 2−9 −3 , ≠ 3 4 , = 3 . Determine whether →3 () exist or does not exist. Solution : →3− () = →3− 2 − 9 − 3 = →3− ( + 3)( − 3) − 3 = →3− ( + 3) = (3 + 3) = 6 →3+ () = →3+ 2 − 9 − 3 = →3+ ( + 3)( − 3) − 3 = →3+ ( + 3) = (3 + 3) = 6 →3− () = →3+ () = 6 ℎ, →3 () = 6 () →3− () = →3− 2 − 9 − 3 = →3− ( + 3)( − 3) − 3 = →3− ( + 3) = (3 + 3) = 6 →3+ () = →3+ 2 − 9 − 3 = →3+ ( + 3)( − 3) − 3 = →3+ ( + 3) = (3 + 3) = 6 →3− () = →3+ () = 6 ℎ, →3 () = 6 ()
MAT183/ MAT421 : Calculus I 32 Example 3 : h(x) piece-wise function < > (limit does not exist) Given ℎ() = { 2−9 −3 , ≤ 3 4, > 3 . Determine whether →3 ℎ() exist or does not exist. Solution : Example 4 : p(x) piece-wise function < > (limit exist) Given () = { 2−9 −3 , ≤ 3 2 , > 3 . Determine whether →3 () exist or does not exist. Solution : →3− ℎ() = →3− 2 − 9 − 3 = →3− ( + 3)( − 3) − 3 = →3− ( + 3) = (3 + 3) = 6 →3+ ℎ() = →3+ 4 = 4 →3− ℎ() ≠ →3+ ℎ() ℎ, →3 ℎ() →3− () = →3− 2 − 9 − 3 = →3− ( + 3)( − 3) − 3 = →3− ( + 3) = (3 + 3) = 6 →3+ () = →3+ 2 = 2(3) = 6 →3− () = →3+ () = 6 ℎ, →3 () = 6 ()
Chapter 1 : Functions, Limits & Continuity 33 CONTINUITY Continuity is another far-reaching concept in calculus. A function can either be continuous or discontinuous. One easy way to test for the continuity of a function is to see whether the graph of a function can be traced with a pen without lifting the pen from the paper. For the math that we are doing in precalculus and calculus, a conceptual definition of continuity like this one is probably sufficient, but for higher math, a more technical definition is needed. Using limits, we’ll learn a better and far more precise way of defining continuity as well. With an understanding of the concepts of limits and continuity, you are ready for calculus. What is Continuity ? In Calculus, a function is continuous at x = a if and only if all three of the following conditions are met : If any one of the conditions is false, then we say that f is discontinuous at x = a, or that it has a discontinuity at x = a. There are two types of question involving continuity, test the continuity (use all the three conditions, step by step) determine the value of constant(s). (usually, we use →− () = →+ () 1. () is defined 2. → () exist or →− () = →+ () 3. → () = ()
MAT183/ MAT421 : Calculus I 34 Example 1 : Given ( ) − − + = 3 2 , 2 2 1 , 2 3 x x x x x f x Test whether f(x) is continuous at x = 2. Solution : Condition 1 : f(2) must defined From ( ) 2 1 3 f x = x − x + (2) (2) 2(2) 1 5 3 f = − + = (defined) Condition 2 : f(x) x 2 lim → must exist f(x) x→ − 2 lim = lim ( 2 1) 3 2 − + → − x x x = (2) 2(2) 1 5 3 − + = f(x) x→ + 2 lim = lim (3 2) 3(2) 2 4 2 − = − = → + x x Since f(x) x→ − 2 lim ≠ f(x) x→ + 2 lim , f(x) x 2 lim → does not exist. So, no need to proceed with the third condition. And since one of the conditions is fail, thus the conclusion is, The function is discontinuous at x = 2. Extra notes : Additional explanation using graphic representation : the graph is disconnected at x = 2
Chapter 1 : Functions, Limits & Continuity 35 Example 2 : Given ( ) + − + − = 1 , 1 , 1 2 x x x x x g x Test whether g(x) is continuous at x = −1. Solution : Condition 1 : g(− 1) must defined From g(x) = x + 1 g(−1) = −1+ 1 = 0 (defined) Condition 2 : g(x) x 1 lim →− must exist g(x) x − →−1 lim = (x x) x + − →− 2 1 lim = ( 1) ( 1) 0 2 − + − = g(x) x + →−1 lim = lim ( 1) 1 1 0 1 + = − + = + →− x x Since g(x) x − →−1 lim = g(x) x + →−1 lim = 0, g(x) x 1 lim →− = 0 (exist) So, we need to proceed with the third condition. Condition 3 : lim ( ) ( 1) 1 = − →− g x g x must exist lim ( ) ( 1) 0 1 = − = →− g x g x Since all the three conditions are met, thus the conclusion is, The function is continuous at x = −1.
MAT183/ MAT421 : Calculus I 36 Example 3 : Given ( ) + = 11, 4 2 3 , 3 4 , 3 2 x x x x x f x Test whether f(x) is continuous at everywhere. Solution : test the continuity at x = 3 Condition 1 : f(3) must defined From f(x) = 2x + 3 f(3) = 2(3) + 3 = 9 (defined) Condition 2 : f(x) x 3 lim → must exist f(x) x→ − 3 lim = 2 3 lim x x→ − = (3) 9 2 = f(x) x→ + 3 lim = lim (2 3) 2(3) 3 9 3 + = + = → + x x Since f(x) x→ − 3 lim = f(x) x→ + 3 lim = 9, f(x) x 3 lim → = 9 (exist) So, we need to proceed with the third condition. Condition 3 : lim ( ) (3) 3 f x f x = → must exist lim ( ) (3) 9 3 = = → f x f x Since all the three conditions are met, thus the conclusion is, The function is continuous at x = 3. Notes: • Continuous at everywhere means we test continuity on x = 3 and x = 4 (boundary values) • If the function continuous at all values of x, it says the function is continuous at everywhere. • But, If the function is not continuous on one of the x values, it says the function is discontinuous at everywhere.
Chapter 1 : Functions, Limits & Continuity 37 test the continuity at x = 4 Condition 1 : f(4) must defined From f(x) = 11 f(4) = 11 (defined) Condition 2 : f(x) x 4 lim → must exist f(x) x→ − 4 lim = lim 2 3 2(4) 3 11 4 + = + = → − x x f(x) x→ + 4 lim = lim 11 11 4 = → + x Since f(x) x→ − 4 lim = f(x) x→ + 4 lim = 11, f(x) x 4 lim → = 11 (exist) So, we need to proceed with the third condition. Condition 3 : lim ( ) (4) 4 f x f x = → must exist lim ( ) (4) 11 4 = = → f x f x Since all the three conditions are met, thus the conclusion is, The function is continuous at x = 4. Grand conclusion : Since the function continuous for all values of x (including the boundary value x = 3 and x = 4), thus the function is continuous at everywhere.
MAT183/ MAT421 : Calculus I 38 Example 4 : Given ( ) + + − + − = 28 , 3 2 1 , 1 3 5 7 , 1 2 x x x x x x h x Test whether h(x) is continuous at everywhere. Solution : test the continuity at x = −1 Condition 1 : h(− 1) must defined Use ( ) 2 1 2 h x = x + x + ( 1) 2( 1) ( 1) 1 2 2 h − = − + − + = (defined) Condition 2 : h(x) x 1 lim →− must exist h(x) x − →−1 lim = lim (5 7) 5( 1) 7 2 1 + = − + = − →− x x h(x) x + →−1 lim = lim (2 1) 2( 1) ( 1) 1 2 2 2 1 + + = − + − + = + →− x x x Since h(x) x − →−1 lim = h(x) x + →−1 lim = 2, h(x) x 1 lim →− = 2 (exist) So, we need to proceed with the third condition. Condition 3 : lim ( ) ( 1) 1 = − →− h x h x must exist lim ( ) ( 1) 2 1 = − = →− h x h x Since all the three conditions are met, thus the conclusion is, The function is continuous at x = −1.
Chapter 1 : Functions, Limits & Continuity 39 test the continuity at x = 3 Condition 1 : h(3) must defined Use ( ) 2 1 2 h x = x + x + (3) 2(3) (3) 1 22 2 h = + + = (defined) Condition 2 : f(x) x 3 lim → must exist h(x) x→ − 3 lim = lim (2 1) 2(3) (3) 1 22 2 2 3 + + = + + = → − x x x f(x) x→ + 3 lim = lim 28 28 3 = → + x Since h(x) x→ − 3 lim ≠ h(x) x→ + 3 lim , h(x) x 3 lim → does not exist So, no need to proceed with the third condition. And since one of the conditions is fail, thus the conclusion is, The function is discontinuous at x = 3. Grand conclusion : Since the function is not continuous on one of the x values, it says the function is discontinuous at everywhere.
MAT183/ MAT421 : Calculus I 40 Example 5/ Page 170/ MAR 2016/ MAT183/ Q1b/ 9 marks The function f(x) is defined as follows. () = { + 1, ≤ 2 2 − 3, 2 < < 4 3 2 2 + 5, ≥ 4 i) Compute the value of k so that →2 () exists. ii) Hence, determine whether f is continuous at x = 4. Solution : i) Compute the value of k so that f(x) x 2 lim → exists. f(x) x 2 lim → exists if and only if f(x) f(x) x x → − → + = 2 2 lim lim lim ( 1) lim ( 3) 2 2 2 + = − → − → + kx kx x x Using direct substitution (no more limits) (2) 1 (2) 3 2 k + = k − 2k +1 = 4k − 3 3 +1 = 4k − 2k 2k = 4 k = 2 # ii) Hence, determine whether f is continuous at x = 4. Note : • To find the constant k, we can jump to condition 2. • No need to follow all the three condition Note : • To test the continuity , we need to follow all the three condition Try This !
Chapter 1 : Functions, Limits & Continuity 41 Example 6/ Page 154/ SEP 2014/ MAT183/ Q1b/ 7 marks The function g(x) is defined as follows. () = { + 1 , < 0 2 2 , 0 ≤ ≤ 2 −2 + , > 2 i) Determine whether g(x) is continuous at x = 0. ii) Find the value of k such that g(x) is continuous at x = 2. Solution : Try This !
MAT183/ MAT421 : Calculus I 42 Example 7/ Page 53/ OCT 2008/ MAT183/ Q1c/ 9 marks The function f(x) is defined as follows, () = { − 1, ≤ 1 3 + , 1 < ≤ 4 2−5+4 −4 , > 4 Given that →1 () = 4, i) find the values of a and b. ii) determine whether f(x) is continuous at x = 4. Solution : i) find the values of a and b. Given that →1 () = 4, →1− () = →1+ () = 4 where →1− () = 4 →1− ( − 1) = 4 Using direct substitution (no more limits) (1) − 1 = 4 = 5 # ii) determine whether f(x) is continuous at x = 4. and →1+ () = 4 →1+ (3 + ) = 4 Using direct substitution (no more limits) 3(1) + = 4 = 1 # Try This !
Chapter 1 : Functions, Limits & Continuity 43 Example 8/ APR 2008/ MAT183/ Q1b/ 5 marks find the values of m and n if h(x) is continuous at x = 0. ℎ() = { , < 0 , = 0 + 5, > 0 Solution : We jump to the second condition, h(x) is continuous at x = 0 if h(x) x 0 lim → exist thus, h(x) h(x) x x → − → + = 0 0 lim lim where, ( ) x mx h x x x sin lim lim 0 0 → − → − = = mx mx m x sin lim 0 → − = m(1) = m And ( ) = → + h x x 0 lim lim ( 5) 0 + → + x x = 0 + 5 (using direct substitution, no more limits) = 5 from h(x) h(x) x x → − → + = 0 0 lim lim m = 5 # To find n, we jump to the third condition, h(x) is continuous at x = 0 if lim ( ) (0) 0 h x h x = → thus, lim ( ) (0) 0 h x h x = → − n x mx x = → − sin lim 0 m = n n = 5 # END OF CHAPTER 1 use h(x) = n h(0) = n (using Squeezing theorem, adjust the denominator so that it is equal to the angle of sine )
MAT183/ MAT421 : Calculus I 44 Tutorial 1.1 : LIMITS 1. Evaluate the following limits. ) →+∞ √5 2 + 2 − 3 ) →1 √2 + 1 − √3 − 1 ) →0 2 − 2 3 2 2. Evaluate the following limits. ) →+∞ √ 2 + 4 6 − 5 ) →7 √3 − 5 − 4 − 7 ) →0 4 − 4 2 3 3. Evaluate each of the following limits. ) →3 2 − 9 2 2 − 7 + 3 ) →∞ 5 − 3 √ 2 + 1 ) →0+ ( 2√ ) 4. Evaluate the following limits. ) →0 8 2 2 ) →4 √ + 5 − 3 2 − 8 ) →∞ 2 − 7 + 10 − 5 5. Find ) →∞ 2 3 − 5 ) →0 2 3 ) →2 √4 − √6 − 2 − 4 6. Evaluate each of the following limits. ) →3 − 3 3 − 2 − 6 ) →0 ( 2 ) ) →2 2 − √ + 2 − √3 − 2 7. Evaluate each of the following limits. ) →2 4 − 2 − 2 ) →−1 √ 2 + 8 − 3 + 1 ) →0 2 8. Evaluate the following limits. ) →1 2 − 1 − 1 ) →2 − 2 2 − √6 − ) →∞ 2 − 1 2 + 3 + 5 Ans : 8. ) 2 ) 4 ) 0 Ans : 7. ) − 1 ) − 1 3 ) 2 Ans : 6. ) 1 15 ) 1 2 ) 2 Ans : 5. ) 0 ) 2 3 ) 1 16 Ans : 2. ) 1 6 ) 3 8 ) − 4 Ans : 1. ) √5 ) 1 √3 ) 0 Ans : 3. ) 6 5 ) 5 ) 0 Ans : 4. ) 4 ) 1 12 ) ∞
Chapter 1 : Functions, Limits & Continuity 45 Tutorial 1. 2 : CONTINUITY 1. Sep 2011/MAT183/Q1b 2. Jun 2016/MAT421/Q1c 3. Apr 2010/MAT183/Q1b Answer : 3. i) k = −7 ii) f(x) is discontinuous at x = 2 iii) −∞ Answer : 1. = 4, n= −2 g(x) is discontinuous at x = 3 Answer : 2. = 12 5 , b= −8