   Chapter 2.6, Problem 51E ### 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 = x 3 − x x 2 − 6 x + 5

To determine

To find: The horizontal and vertical asymptotes of y=x3xx26x+5.

Explanation

Theorem used:

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

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=x3xx26x+5.

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)=x3xx26x+5.

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

f(x)=x3xx2x26x+5x2=x3x2xx2x2x26xx2+5x2=x1x16x+5x2

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

limxx3xx26x+5limxx1x16x+5x2

As x goes to infinity, x1x goes to infinity. That is,

limx(x1x)=limx(x)limx(1x)=[by theorem1]

As x goes to negative infinity, 1+1x4 goes to 1. That is,

limx(16x+5x2)=limx(1)limx(6x)+limx(5x2)=10+0[by theorem1]=1

Therefore, f(x) approaches to large positive infinity as x goes to large infinity. That is,

limx(x3xx26x+5)=.

Since the limit of the function does not approach any constant value y=L, there is no horizontal asymptotes.

Thus, the function does not have a horizontal asymptote.

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

limxx3xx26x+5limxx1x16x+5x2

As x goes to negative infinity, x1x goes to negative infinity. That is,

limx(x1x)=limx(x)limx(1x)=[by theorem2]

As x goes to negative infinity, 1+1x4 goes to 1

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