   Chapter 2.3, Problem 22E ### 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 u → 2 4 u + 1 − 3 u − 2

To determine

To evaluate: The limit of the function limu24u+13u2.

Explanation

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).

Difference of square formula: (a2b2)=(a+b)(ab)

Fact 1:

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

Evaluation:

Let f(u)=4u+13u2 (1)

Note 1:

The direct substitution method is not applicable for the function f(u) since the function f(0) is indeterminate form when u=2. That is,

f(2)=4(2)+13(2)2=930=3322=00

Note 2:

The Quotient rule is not applicable for the function f(u) as the limit of the denominator is zero. That is,

limu2(u2)=limu2(u)limu22 (by limit law 2)=(2)(2) (by limit law 8 and 7)=0

Note 3:

The limit may be infinite or it may be some finite value when both the numerator and the denominator approach 0.”

By note 3, take the limit u approaches 2 but u2.

Simplify f(u) by using elementary algebra.

f(u)=4u+13u2

Take the conjugate of the numerator and multiply and divide of f(u)

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