   Chapter 2.3, Problem 11E ### 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

# Evaluate the limit, if it exists. lim x → 2 x 2 + x − 5 x − 5

To determine

To evaluate: The limit of the function limx5x26x+5x5.

Explanation

Limit Laws:

Suppose that c is a constant and the limits limxaf(x) and limxag(x) exists, then

Limit law 2: limxa[f(x)g(x)]=limxaf(x)limxag(x)

Limit law 7: limxac=c

Limit law 8: limxax=a

Direct substitution property:

If f is a polynomial or a rational function and a is in the domain of f, then limxaf(x)=f(a).

Fact 1:

If f(x)=g(x) when xa, then limxaf(x)=limxag(x), provided the limit exist.

Calculation:

Compute the limit value of the denominator.

limx5(x5)=limx5(x)limx5(5) (by limit law 2)=limx5(x)5 (by limit law 7)=55 (by limit law 8)=0

Since the limit of the denominator is zero, the quotient law cannot be applied

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