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Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

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BuyFindarrow_forward

Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343
Textbook Problem

Estimate the horizontal asymptote of the function

f ( x ) = 3 x 3 + 500 x 2 x 3 + 500 x 2 + 100 x + 2000

by graphing .f for –10 ≤ x ≤ 10. Then calculate the equation of the asymptote by evaluating the limit. How do you explain the discrepancy?

To determine

To estimate: The horizontal asymptote of f(x)=3x3+500x2x3+500x2+100x+2000 by graphing f for 10x10; then find the exact equation of the asymptote by evaluating the limit and explain the discrepancy of the graph.

Explanation

Use the online graphing calculator and draw the graph of f(x)=3x3+500x2x3+500x2+100x+2000 as shown below in Figure 1.

From Figure 1, it is concluded that the equation of the horizontal asymptote is y=1.

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

Limit law 9: limxaxn=an, where n is a positive integer.

Limit law 10: limxaxn=an, where n is a positive integer, if n is even, assume that a>0.

Limit law 11: limxaf(x)n=limxaf(x)n, where n is a positive integer, if n is even, assume that limxaf(x)>0.

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:

Obtain the value of the function as x approaches infinity.

Consider f(x)=3x3+500x2x3+500x2+100x+2000.

Divide both the numerator and the denominator by the highest power of x in the denominator.

f(x)=3x3+500x2x3(x3+500x2+100x+2000x3)=3x3x3+500x2x3(x3x3+500x2x3+100xx3+2000x3)=3+500x(1+500x+100x2+2000x3)

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

limxf(x)=limx(3+500x(1+500x+100x2+2000x3))=limx(3+500x)limx(1+500x+100x2+2000x3)[by limit law 5]=limx

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