   Chapter 2.6, Problem 23E ### 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 limit or show that it does not exist. lim x → − ∞ 1 + 4 x 6 2 − x 3

To determine

To find: The value of limx1+4x62x3.

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

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

Calculation:

Obtain the value of the function as x approaches infinity.

Consider the function f(x)=1+4x62x3.

f(x)=1+4x62x3          =x6(1x6+4)2x3         =x6(1x6+4)2x3         =(x6)12(1x6+4)2x3

Simplify further f(x) by using elementary algebra,

f(x)=(x2)32(1x6+4)2x3=(x2)3(1x6+4)2x3=(|x|)3(1x6+4)2x3[|x|=x2]

Since x>0, |x|=x

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