   Chapter 2.3, Problem 64E ### 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 lim x → 2 6 − x − 2 3 − x − 1 .

To determine

To evaluate: The limit of the function limx26x23x1.

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(x)=6x23x1 (1)

Note 1:

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

f(2)=6x23x1=622321=4211=00

Note 2:

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

limx2(3x1)=limx2(3x)limx21(by limit law 2)=limx2(3x)(1)(by limit law 11 and 3)=(32)1=11

=11=0

Note 3:

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

Calculation:

By note 3, take the limit u approaches 2 but x2.

Simplify f(x) by using elementary algebra.

f(x)=6x23x1

Take the conjugate of the numerator and denominator multiply and divide of f(x)

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