   Chapter 2.3, Problem 15E ### 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 t → − 3 t 2 − 9 2 t 2 + 7 t + 3

To determine

To evaluate: The limit of the function limt3t292t2+7t+3.

Explanation

Limit Laws:

Suppose that c is a constant and the limits limxaf(x) and limxag(x) exist, 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 7:limxac=c

Limit law 8:limxax=a

Limit law 9:limxaxn=an,  where n is a positive integer.

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:

The limit of the denominator is zero.

limt3(2t2+7t+3)=limt3(2t2)+limt3(7t)+limt3(3) (by limit law 1)=2limt3(t2)+7limt3(t)+limt3(3) (by limit law 3)=2(3)2+7(3)+3 (by limit law 9,8,7)=0

The quotient law cannot be used as the denominator of the function tends to zero when t3

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