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Math
CalculusSingle Variable Calculus: Early Transcendentals, Volume I8th Edition

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Chapter 2.3, Problem 20E
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Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

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BuyFindarrow_forward

Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343
Textbook Problem

Evaluate the limit, if it exists.

lim t → 1 t 4 − 1 t 3 − 1

To determine

To evaluate: The limit of the function limt→1t4−1t3−1.

Explanation

Direct substitution property:

If f is a polynomial or a rational function and a is in the domain of f, then limx→af(x)=f(a).

Difference of cube formula: (a3−b3)=(a−b)(a2+ab+b2)

Difference of square formula: (a2−b2)=(a+b)(a−b)

Fact 1:

If f(x)=g(x) when x≠a, then limx→af(x)=limx→ag(x), provided the limit exist.

Let f(t)=t4−1t3−1. (1)

Note 1:

The direct substitution method is not applicable for the function f(t) since the function f(0) is in indeterminate form at t=1.

f(1)=(1)4−1(1)3−1=1−11−100

Note 2:

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

limt→1(t3−1)=limt→1(t3)−limt→11          (by limit law 2)=(1)3−1 (by limit law 9 and 7)=1−1=0

Note 3:

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

Calculation:

By note 3, take the limit t approaches 1 but t≠1.

Simplify f(t) by using elementary algebra

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Sect-2.1 P-1ESect-2.1 P-2ESect-2.1 P-3ESect-2.1 P-4ESect-2.1 P-5ESect-2.1 P-6ESect-2.1 P-7ESect-2.1 P-8ESect-2.1 P-9ESect-2.2 P-1E
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