   Chapter 2.6, Problem 52E ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

#### Solutions

Chapter
Section ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343
Textbook Problem

# Find the horizontal and vertical asymptotes of each curve. If you have a graphing device, check: your work by graphing the curve and estimating the asymptotes. y = 2 e x e x − 5

To determine

To find: The horizontal and vertical asymptotes of y=2exex5.

Explanation

Definition used:

Limit Laws used: Suppose that c is a constant and the limits limxaf(x) and limxag(x) exists, then

Limit law 1: limxa[f(x)+g(x)]=limxaf(x)+limxag(x)

Limit law 2: limxa[f(x)g(x)]=limxaf(x)limxag(x)

Limit law 3: limxa[cf(x)]=climxaf(x)

Limit law 4: limxa[f(x)g(x)]=limxaf(x)limxag(x)

Limit law 5: limxaf(x)g(x)=limxaf(x)limxag(x) if limxag(x)0

Limit law 6: limxa[f(x)]n=[limxaf(x)]n where n is a positive integer.

Limit law 7: limxac=c

Definition used: Horizontal Asymptote:

The line y=L is called a horizontal asymptote of the curve y=f(x) if either limxf(x)=L and limxf(x)=L.

Calculation:

Consider the function, f(x)=2exex5.

Compute the limit of f(x) as x approaches infinity.

limx2exex5limx2exex5×exex=limx(2e0ex(ex5))=limx(2(1)e05ex)=limx(215ex)

Apply the appropriate laws and simplify further.

limx2exex5=limx(2)limx(1)limx(5ex)[by limit law 1,2]=215limx(ex)[by limit law 3,7]=215(0)=2

Therefore, the function f(x)=2exex5 is approach 2 as x approaches infinity. That is,

limx2exex5=2.

Since, limx2exex5=2 the line y=2 is called horizontal asymptote of the curve y=2exex5.

Thus, the horizontal asymptote is x=2

Compute the limit of f(x) as x approaches negative infinity

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Study Guide for Stewart's Single Variable Calculus: Early Transcendentals, 8th 