   Chapter 2.6, Problem 30E ### 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 → ∞ ( x 2 + 1 )

To determine

To show: The value of limxx2+1 does not exist.

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.

Proof:

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

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

limxx2+1=limxx2(1+1x2)=limx(x21+1x2)=limx(x1+1x2)[x2=x for x>0]

The limit law limxaf(x)g(x)=limxaf(x)limxag(x) is valid for one sided limits

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