2. Mat 121-Limits.pdf

www.covenantuniversity.edu.ng Raising a new Generation of Leaders OPEN DISTANT LEARNING CENTRE COURSE MATRIAL MAT121: Calculus III MODULE 2: LIMIT OF FUNCTIONS...

www.covenantuniversity.edu.ng Raising a new Generation of Leaders OPEN DISTANT LEARNING CENTRE COURSE MATRIAL MAT121: Calculus III MODULE 2: LIMIT OF FUNCTIONS 2  UNIT 1: LIMIT OF FUNCTIONS Unit Objectives After studying this unit, you should be able to: ❖define a limit; ❖evaluate the limit of a function by substitution; ❖evaluate the limit of a function by dividing out technique; ❖evaluate the limit of a function by rationalization; ❖compute one-sided limits; ❖determine if the limit of a function exists 3  DEFINITION OF LIMIT OF FUNCTIONS x2 - 9 Let us consider the behaviour of the function f(x)= x-3 around a point x=3. X 2.9 2.99 2.999 2.9999 3.1 3.01 3.001 3.0001 f(x) 5.9 5.99 5.999 5.9999 6.1 6.01 6.001 6.0001 You will notice in the Table, that as the values of x moves closer to 3 from both directions, the values of the function f(x) moves closer to 6. This means that the limit of the function f(x) as x approaches 3 is 6. This is written as lim x →3 f(x) = 6. 4 Definition: If the values of f(x) can be made as close as we like to a unique number L by taking values of x sufficiently close to a (but not equal to a), then we write lim f ( x ) = L x →a which is read as “the limit of f(x) as x approaches a is L.” 5  Evaluating Limits Using Basic Limit Results 1. Basic Limit Result: For any real number a and constant c, i) xlim →a x =a ii) xlim →a c=c Example 2.1. Evaluate i) lim x →4 x ii) lim10 x →3 Solution: Using i) you obtain lim x →4 x=4. ii) Using (ii) you obtain lim10 x →3 = 10 6  Evaluating Limits Using Laws of Limits Let b and a be real numbers, let n be a positive integer, and let f and g be functions with the following limits xlim →a f(x) = L and xlim →a g(x) = K , then i. Scalar multiple Law: xlim →a [bf(x)] = bL ii. Sum Law: xlim →a [f(x) + g(x)] = L + K iii. Difference Law: xlim →a [f(x) − g(x)] = L − K iv. Product Law: xlim →a [f(x)g(x)] = LK 7 7  f(x) L v. Quotient Law: lim = x →a g(x) K vi. Power Law: lim[f(x)]n = Ln x →a n vii. Root Law: xlim →a n f(x) = n lim f(x) = L x →a for all L if n is odd and for L  0 if n is even and f(x)  0. 8  2x 2 − 3x + 1 Evaluate i)xlim →−5 (3x + 3) ii) lim x →2 x3 + 4 Solution. i) xlim →−5 (3x + 2) = lim 3x + lim 2 x →−5 x →−5 Applying ii = 3  lim x + lim 2 Applying i x →−5 x →−5 = 3  (−5) + 2 = −17 2 2x − 3x + 1 ( lim 2x 2 − 3x + 1 ) ii) lim = x →2 Applying v x →2 3 x +4 ( lim x + 4 x →2 3 ) 2  lim x 2 − 3  lim x + lim1 x →2 x →2 x →2 = lim x 3 + lim 4 x →2 x →2 ( ) 2 2  lim x − 3  lim x + lim1 2(4) − 3(2) + 1 1 = x →2 x →2 x →2 = = ( lim x ) (8) + 4 3 + lim 4 4 x →2 x →2 9 9  Computation of Limits by Substitution Method x2 + x −6 Evaluate lim x →−1 x +3 x 2 + x − 6 ( −1)2 + ( −1) − 6 Solution. lim = x →−1 x +3 ( −1) + 3 1 −1 − 6 = −1 + 3 −6 = = −3 2 10 Computation of Limits by Dividing through Technique Example: Evaluate lim − 6 x + 9 2 x x →3 x −3 Solution. x 2 − 6x + 9 ( x − 3)2 Factorize: lxim = lim = lim( x − 3) →3 x −3 x →3 x −3 x →3 Substitute 3 for x: lim( x − 3) = 3 − 3 = 0 x →3 11  Computation of Limits by Rationalization Technique x + 1 −1 Example: Evaluate lim x →0 x  x + 1 −1 x + 1 + 1 x + 1 −1 Solution. lim    = lim x →0  x ( x + 1 + 1 x →0 x x + 1 + 1 ) x = lim x →0 x ( ) x +1 +1 1 = lim x →0 x +1 +1 Substituting 0 for x: 1 1 1 1 lim = = = x →0 x +1 +1 0 +1 +1 1 +1 2 12 Evaluating Limits by Simplifying Complex Functions 3 3 − Evaluate lim x + 1 2. x →1 x − 1 Solution. Simplifying yields 3 3 6 − 3(x + 1) 6 − 3x − 3 − 2(x + 1) 2(x + 1) lim x + 1 2 = lim = lim x →1 x − 1 x →1 x −1 x →1 x −1 3 − 3x 2(x + 1) 3 − 3x −3(x − 1) = lim = lim = lim x →1 x − 1 x →1 2(x + 1)(x − 1) x →1 2(x + 1)(x − 1) 13 You now apply the dividing through technique to obtain −3(x − 1) −3 lim = lim x →1 2(x + 1)(x − 1) x →1 2(x + 1) Finally, you use the substitution method −3 −3 3 lim = =− x →1 2(x + 1) 2(1+ 1) 4 14 Evaluating Limits When Laws of Limit Do Not Apply 2 10  Evaluate lim  + x →0  x x(x − 5)  Solution. First simplify the function 2 10  2x(x − 5) + 10x 2x 2 − 10x + 10x 2 lim  + = lim = lim = lim x →0  x x(x − 5)  x →0 x 2(x − 5) x →0 x 2(x − 5) x →0 x − 5 Finally, you apply the substitution technique 2 10  2 2 lim  + = lim = − x →0  x x(x − 5)  x →0 x − 5 5 15  One-Sided Limits: Right Hand Limit: If the values of f(x) can be made close as we like to L by taking values of x sufficiently close to a (but greater than a), then we write lim+ f ( x ) = L. x →a Left Hand Limit: if the values of f(x) can be made close as we like to L by taking values of x sufficiently close to a (but less than a), then we write lim− f ( x ) = L. x →a 16 Evaluate i) xlim →2− |x − 2| ii) lim+ |x − 2| x →2 Solution. i) When you approach x from the left of 2, you have x  0. Subtracting 2 from both sides of the inequality gives x − 2  0. This means |x − 2| equals −(x − 2) for x  2. You use substitution to evaluate this limit lim− |x − 2|= lim− -(x − 2) = −(2 − 2) = 0 x →2 x →2 ii) Approaching x from the right of 2 gives x>0. Subtracting 2 from both sides gives x-2>0. This means |x − 2| equals x-2 for x