LIMITS AND DERIVATIVES CONDITIONS FOR THE EXISTENCE OF A LIMIT

Transcription

1 LIMITS AND DERIVATIVES Te limit of a function is defined as te value of y tat te curve approaces, as x approaces a particular value. Te limit of f (x) as x approaces a is written as f (x) approaces, as x gets closer and closer to a. xa f (x), and represents te value tat A limit will exist at CONDITIONS FOR THE EXISTENCE OF A LIMIT x a if: f is continuous at x a. If tere is point discontinuity at x a. If te function x A limit will not exist at x a if: Tere is discontinuity across an interval and x a lies witin tis interval. A limit does not exist at tis value of x A limit does not exist at tis value of x Te Scool For Excellence 011 Te Essentials Year 11 Matematical Metods Book 1 Page 1

2 EVALUATING LIMITS GRAPHICALLY SINGLE FUNCTIONS Step 1: Observe te value of y as x approaces a value below te point of interest. Step : Observe te value of y as x approaces a value above te point of interest. Step 3: If te left and limit is equal to te rigt and limit a limit exists at x a. Te limit is simply equal to te value of y tat te curve is approacing on eiter side of te point of interest. QUESTION 1 Te grap of f ( x ) is given on side. lim f x is equal to x1.5 A 0 B 3 C 3 D 1.5 E Undefined Solution y x Te Scool For Excellence 011 Te Essentials Year 11 Matematical Metods Book 1 Page

3 LIMITS OF HYBRID FUNCTIONS A ybrid function is a function tat as different rules describing te different sections of its domain. its of ybrid functions are evaluated in te same manner as previously described. Step 1: Find te limit of eac individual function at te given value of x. Step : If te limiting values are te same, te limit exists at tat value of x. If one or more of te limiting values are different, te limit does not exist at tat particular value of x. Note: If te function is discontinuous across an interval and te value of x lies in tis interval, te limit cannot be evaluated QUESTION Te grap of f (x) is sown below. (a) Find lim f ( x) x5 (b) Find lim f ( x) x (c) lim f ( x) lim f( x) x1 x3.5 Te Scool For Excellence 011 Te Essentials Year 11 Matematical Metods Book 1 Page 3

4 EVALUATING LIMITS ALGEBRAICALLY Does te given equation contain a denominator bearing terms involving x? No Yes Substitute te limit value into te given equation and solve. Substitute te limit value into te denominator. If te denominator is equal to zero, factorise and eliminate terms by cancellation. Ten substitute te limit value into te simplified equation and solve. If te denominator is not equal to zero, substitute te limit value into te given equation and solve. Te manner in wic we evaluate limits algebraically depends on weter te denominator of te function carries terms involving x : Given f (x) : xa (a) If te equation does not carry a denominator, or if te denominator as NO terms involving x, simply substitute te value of a into te given equation and simplify. 3 For example: lim(5 x x ) 5 (1) (1) x1 3 (b) If te denominator carries terms involving x, we need to first determine weter te value of x (or a ) in question causes te function to be undefined. Substitute te value of a into te denominator of te equation. If te answer does not equal zero, proceed to substitute te value of a into te equation. For example: x 1 ( ) lim, x x x ( ) 4 4 If te answer is equal to zero, we need to eliminate te term(s) tat is/are causing te function to become undefined. Factorise te given equation and eliminate tese terms by cancellation. Before evaluating te limit, substitute te value of a into te new equation to ensure tere are no oter terms present tat will make te function undefined. For example: lim x3 ( x 5x 6) ( x 3) ( x3)( x) lim lim ( x ) 3 x3 ( x 3) 3 1 x Te Scool For Excellence 011 Te Essentials Year 11 Matematical Metods Book 1 Page 4

5 Note: Wen tere are terms involving x in te denominator of a fraction, tere may be some values of x wic cause te function to become undefined. Tese values of x must be stated wit your answer. To determine wic values of x will make te function undefined, before te function is factorised and terms are eliminated by cancellation: Step 1: Let te bottom of fraction equal zero. Step : Solve for x. For Example: x 3x 4 x 1 ( x 1)( x 3 ) Restrictions: x 1 or x 3. Note: Even toug tis function is undefined at x 1, we may still investigate wat appens to te value of y as x approaces tis value (i.e. te limit). Te Scool For Excellence 011 Te Essentials Year 11 Matematical Metods Book 1 Page 5

6 LIMIT THEOREMS (a) Te limit of a constant is equal to te value of te constant: If f ( x) k ten f ( x) k. xa For Example: 5 5 x (b) Te limit of te sum and/or difference of a series of terms is equal to te sum and/or difference of te limits of eac individual term. xa f x gx f x gx xa xa x For Example: x x x x x x x (c) Te limit of te product of two functions is equal to te product of te limits of eac individual function. xa f x. gx f x gx xa xa x For Example: x 1 x x x 1 x 1 x (d) Te limit of te quotient of two functions is equal to te quotient of te limits of eac individual function. f g x x For Example: f g x x x3 x g x. Note tat 0 x 5 9x 5 x3 x 5 1 x3 x 9x Te Scool For Excellence 011 Te Essentials Year 11 Matematical Metods Book 1 Page 6

7 QUESTION 3 Evaluate lim x5 x x 5, if it exists. x15 Solution Substitute x = -5 into denominator to determine if te function is defined at tis point: Denominator = 0 Factorise and eliminate terms by cancellation: x 5 ( x5)( x5) ( x5) lim lim lim x x15 ( x3)( x5) ( x3) x5 x5 x5 Evaluate te required limit - Substitute x = 3 into te simplified expression: ( x 5) 5 x lim x5 ( 3) 4 State any restrictions on te values of x: x 30 x 3 QUESTION 4 3 lim 14 6t t, if it exists. Evaluate Solution t Te Scool For Excellence 011 Te Essentials Year 11 Matematical Metods Book 1 Page 7

8 QUESTION 5 Evaluate Solution x lim x x 1. QUESTION 6 lim 0 (3 ) 9 is equal to: A 0 B undefined C 6 D 6 E QUESTION 7 Sow tat lim z 6 lim z lim 6 z0 z0 z0. Solution Te Scool For Excellence 011 Te Essentials Year 11 Matematical Metods Book 1 Page 8

9 QUESTION 8 Given lim f( x) 9, x8 lim gx ( ) and x8 lim x ( ) 4, determine te following limits. x8 (a) lim f ( x) 1 ( x) x8 (b) lim gxx ( ). ( ) f( x) x8 (c) lim f ( x) g( x) ( x) x8 Te Scool For Excellence 011 Te Essentials Year 11 Matematical Metods Book 1 Page 9

10 DIFFERENTIATION Te derivative describes te gradient of te tangent to a curve at any value of x. Te process of finding a derivative is called differentiation. Tangent Te gradient of a line is constant and does not cange at different values of x. As te derivative represents te gradient to a curve, te derivative of a linear expressions will result in a constant (numeric) value. Te gradient to a curve modelled by a non linear equation canges, and is dependent on te value of x. Terefore, te derivative of a non linear expression will result in an expression in terms of x. Only wen a particular value of x is substituted into te derivative equation, will a numeric value be obtained. Tis value represents te gradient of te tangent to te curve at te specific value of x. Wen te equation is denoted as f (x), te derivative is represented as f '( x). dy Wen te equation is denoted as y, te derivative is represented as. dx dy Note: dx is an operation - it is not a quotient (it is not te same as dy dx). Note: d dx reads te derivative of wit respect to x. Derivatives may be evaluated from first principals or by using a set of rules. Wenever you see te following words/prase: gradient, gradient function, gradient of te tangent, automatically tink of differentiation. Te Scool For Excellence 011 Te Essentials Year 11 Matematical Metods Book 1 Page 10

11 DERIVATIVES FROM FIRST PRINCIPLES We can use limits to find derivatives using a process tat is referred to as differentiation using first principles. Te derivative from first principles is obtained by applying te rule: Derivative f '( x) ( y) f ( x) 0 [ f( x) f( x)] dy d d dx dx dx An alternative notation for te limit teorem is: [ ( ) ( )] lim f x x f x x δx0 were x is a small increment (cange) in x. METHOD: FINDING THE DERIVATIVE FROM FIRST PRINCIPLES Step 1: Write te expressions for f (x) and f ( x ). Step : Substitute te expressions for f (x) and f ( x ) into te limit teorem. dy dx [ f ( x ) 0 f ( x)] Step 3: Expand and collect like terms. Step 4: Remove as a common factor and simplify. Step 5: Substitute 0. Note: f (x) represents an equation in terms of x. To obtain f ( x ) from f (x), replace x in te given equation wit ( x ). For Example: If f ( x) x 6x 1, ten f ( x ) ( x ) 6( x ) 1. To find te derivative at a specific value of x, substitute te given value of x into te derived equation. Note: Te gradient at x is denoted as f '(). Te Scool For Excellence 011 Te Essentials Year 11 Matematical Metods Book 1 Page 11

12 QUESTION 9 Find te derivative of f ( x) x 3x using first principles. Hence find te gradient of te tangent to te curve at x. Solution Write te expressions for f (x) and f ( x ) : If f ( x) x 3x ten f ( x ) ( x ) 3( x ) Substitute te expressions for f (x) and f ( x ) into te limit teorem: f ( x) f( x) f( x) lim 0 ( x ) 3( x) ( x 3 x) lim 0 Expand and collect like terms: x x 3x 3 x 3x lim 0 x 3 (x 3) lim lim lim(x 3) x 3, Te derivative or te equation describing te gradient to te curve y x 3x is x 3. Substitute in te value of x : Wen x, te gradient is equal to f ( ) 1. NOTE: Te lim 0 notation was included in eac step, until 0 was substituted into te equation. You may ceck your answers by differentiating te given equation by rule. If te two answers are different, it is likely tat an error as been made wen expanding te brackets in te limit formula. Remember to multiply every term in te second set of brackets by negative one! Te Scool For Excellence 011 Te Essentials Year 11 Matematical Metods Book 1 Page 1

13 QUESTION 10 x 1 (a) Sow, using first principles, tat te derivative of f( x) is f '( x). x 1 ( x 1) Te Scool For Excellence 011 Te Essentials Year 11 Matematical Metods Book 1 Page 13

14 (b) Hence find f '( 5). (c) Find te coordinates of te point(s) on te curve at wic te gradient of te tangent equals 4. Te Scool For Excellence 011 Te Essentials Year 11 Matematical Metods Book 1 Page 14