   Chapter 2.6, Problem 17E ### 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 x 2 + 1

To determine

To find: The value of limxx2x2+1.

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

Theorem used: If r>0 is a rational number such that xr is defined for all x, then limx1xr=0.

Calculation:

Obtain the value of the function as x approaches negative infinity.

Consider f(x)=x2x2+1.

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

f(x)=x2x2x2+1x2=xx22x2x2x2+1x2=1x2x21+1x2

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

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