Chapter 2.6, Problem 49E

Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

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 x 2 + x − 1 x 2 + x − 2

To determine

To find: The horizontal and vertical asymptotes of y=2x2+x1x2+x2.

Explanation

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

Limit law 8: limxax=a

Theorem used 1: If r>0 is a rational number, then limx1xr=0.

Theorem used 2: If r>0 is a rational number such that xr is defined, then limx1xr=0.

Calculation:

Horizontal asymptote:

Obtain the horizontal asymptote of the function y=2x2+x1x2+x2.

Recall the definition of 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”.

Consider the function f(x)=2x2+x1x2+x2.

Divide both the numerator and the denominator by the highest power of x in the denominator. That is, x20.

f(x)=2x2+x1x2x2+x2x2=2x2x2+xx21x2x2x2+xx22x2=2+1x1x21+1x2x2

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

limx2x2+x1x2+x2limx2+1x1x21+1x2x2                    =limx(2+1x1x2)limx(1+1x2x2)[by limit law 5]

Apply the appropriate laws and simplify further.

limx2x2+x1x2+x2=limx(2)+limx(1x)limx(1x2)limx(1)+limx(1x)limx(2x2)[by limit law 1,2]=2+limx(1x)limx(1x2)1+limx(1x)2limx(1x2)[by limit law 3,7]

Since r>0, apply the theorem1 stated above to compute the value of the limit function.

limx2x2+x1x2+x2=2+(0)(0)1+(0)2(0)=21=2

Therefore, the function f(x)=2x2+x1x2+x2 is approach 2 as x approaches infinity. That is, limx2x2+x1x2+x2=2.

Since limx2x2+x1x2+x2=2, the line y=2 is called a horizontal asymptote of the curve y=2x2+x1x2+x2.

Thus, the horizontal asymptote is 2.

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

limx2x2+x1x2+x2limx2+1x1x21+1x2x2                    =limx(2+1x1x2)limx(1+1x2x2)[by limit law 5]

Apply the appropriate laws and simplify further

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