Chapter 2.3, Problem 65E

### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

Chapter
Section

### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343
Textbook Problem

# Is there a number a such that lim x → − 2 3 x 2 + a x + a + 3 x 2 + x − 2 exists? If so, find the value of a and the value of the limit

To determine

To find: The value of a and the limit of the function limx23x2+ax+a+3x2+x2.

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)=3x2+ax+a+3x2+x2.

Note 1:

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

limx2x2+x2=limx2(x2+x)limx2(2)(by limit law 2)=limx2x2+limx2xlimx2(2)(by limit law 2)=(2)2+(2)(2)(by limit law 9,8 and 7)=0

Note 2:

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

Calculation:

By note 1, the denominator of the function is zero. By note 3, the limit of the function exists only if the limit of the numerator approaches 0 as x approaches −2.

That is, compute limx2(3x2+ax+a+3)=0

limx2(3x2+ax+a+3)=limx23x2+limx2ax+limx2(a+3)[by limit law 1]=3limx2x2+alimx2x+limx2(a+3)[by limit law 3]=3(2)2+a(2)+(a+3)[by limit law 9,8 and 7]=15a

Equating the limit of the numerator approaches 0 as x approaches −2

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