   Chapter 2, Problem 17RE ### 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. lim x → ∞ ( x 2 + 4 x + 1 − x )

To determine

To find: The value of limx(x2+4x+1x).

Explanation

Result used:

Limit Laws: 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:

Consider the function f(x)=(x2+4x+1x).

Simplify the function as follows.

f(x)=(x2+4x+1x)        =(x2+4x+1x)1×(x2+4x+1+x)(x2+4x+1+x)        =((x2+4x+1)2(x)2)(x2+4x+1+x)        =x2+4x+1x2(x2+4x+1+x)

On further simplification,

f(x)=4x+1(x2+4x+1+x)=4x+1x2(1+4x+1x2)+x=4x+1x2(1+4x+1x2)+x=4x+1|x|(1+4x+1x2)+x[x2=|x|]

Since x>0, then |x|=x

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

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