   Chapter 2.3, Problem 28E ### 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 − 4 x + 4 x 4 − 3 x 2 − 4

To determine

To evaluate: The limit of the function limx2x24x+4x43x24.

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:

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.

f(2)=(2)24(2)+4(2)43(2)24=48+416124=881616=00

Note 2:

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

limx2(x43x24)=limx2x4limx2(3x2)limx2(4) (by limit law 2)=limx2x43limx2(x2)limx2(4) (by limit law 3)=(2)43(2)24 (by limit law 9 and 7)=0

Note 3:

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

Calculation:

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

Simplify f(x) by using elementary algebra.

limx2x24x+4x43x24

Factorise the denominator of f(x).

x43x24=x44x2+x24=x2(x24)+(x24)=(x2+1)(x24)

Factorise the numerator of f(x)

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