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1 (x+a) 1 Limits for e Consider the function g(x) = (1 + )

Calculus: Early Transcendentals | 1st Edition | ISBN: 9780321570567 | Authors: William L. Briggs, Lyle Cochran, Bernard Gillett ISBN: 9780321570567 2

Solution for problem 75RE Chapter 4

Calculus: Early Transcendentals | 1st Edition

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Calculus: Early Transcendentals | 1st Edition | ISBN: 9780321570567 | Authors: William L. Briggs, Lyle Cochran, Bernard Gillett

Calculus: Early Transcendentals | 1st Edition

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Problem 75RE

1 (x+a) 1 Limits for ?e? Consider the function g(x) = (1 + ) x . Show that if 0 ? a < 2 ?,? then g(x) ? efrom ?below? as ?x ?? ?; if 2 ? a < 1 , then ?g?(?x?) ? e? from ? bove? as x? ? ? ?.

Step-by-Step Solution:

Solution 75RE Step 1 1 1(x+a) In this problem we have to show that if 0 a < t2en limx ) x = efrom below and (x+a) if 2 a < 1then lim(1+ ) x = e from above x 1 (x+a) Considerg(x) = (1+ ) x Tak ing limi on both sides we get, 1 (x+a) xg(x) = limx ) x Substitute limit x , we get = (1+0) = 1 Hence we cannot find the limit directly. So first let us split the functions and find limit separately. Then finally we combine to get the result. 1 h(x) Let f(x) = (1+ )axd h(x) = (x+a) then g(x) = (f(x)) Let us find xf(x)and lixx) Consider limf(x) x 1 xf(x) = lx1+ ) x (1+0) = 1 Now consider limh(x) x limh(x) = lim(x+a) = x x h(x) We have g(x) = (f(x)) Taking natural logarithm(ln) on both sides, we get h(x) ln g(x) = ln (f(x)) a ln g(x) = h(x) ln (f(x)) (Since ln x = a ln x) Now taking limit x on both sides, we get xmln g(x) = lix(x) ln f(x) By using limln f(x) = ln (lim f(x))to left hand side we get, x x ln (lim g(x)) = lim h(x) ln f(x) x x Raising to power e on both sides, we get lnxim g(x)) x h(x) ln f(x) e = e lim h(x) limg(x) = e x 1/ln f(Logarithm and e cancel each other in Left hand side) x Since we have f(x) = (1+ )and h(x) = (x+a) we get, x x1/ln (1+ ) = e x …. (1) Step 2 We are given that 0 a < . So let us take a = 0. 2 Therefore (1) becomes lim x+0 = e x1/ln x1+ ) lim 1/ln (1+ ) = e x x ln (1+ ) lim 1x = e x x We have x , then 1 1 = 0. Thus by changing limit from x to 1 0we get x x lim ln1x1+ ) x0 x = e 1 Substituting limit x 0we get = e 1 = e (x+a) Hence if 0 a < then lim(1+ ) 1 = e from below. 2 x x

Step 3 of 3

Chapter 4, Problem 75RE is Solved
Textbook: Calculus: Early Transcendentals
Edition: 1
Author: William L. Briggs, Lyle Cochran, Bernard Gillett
ISBN: 9780321570567

The answer to “1 (x+a) 1 Limits for ?e? Consider the function g(x) = (1 + ) x . Show that if 0 ? a < 2 ?,? then g(x) ? efrom ?below? as ?x ?? ?; if 2 ? a < 1 , then ?g?(?x?) ? e? from ? bove? as x? ? ? ?.” is broken down into a number of easy to follow steps, and 53 words. Calculus: Early Transcendentals was written by and is associated to the ISBN: 9780321570567. This full solution covers the following key subjects: below, bove, consider, efrom, function. This expansive textbook survival guide covers 85 chapters, and 5218 solutions. Since the solution to 75RE from 4 chapter was answered, more than 294 students have viewed the full step-by-step answer. The full step-by-step solution to problem: 75RE from chapter: 4 was answered by , our top Calculus solution expert on 03/03/17, 03:45PM. This textbook survival guide was created for the textbook: Calculus: Early Transcendentals, edition: 1.

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1 (x+a) 1 Limits for e Consider the function g(x) = (1 + )

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