   Chapter 2.3, Problem 20E ### 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 t → 1 t 4 − 1 t 3 − 1

To determine

To evaluate: The limit of the function limt1t41t31.

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 cube formula: (a3b3)=(ab)(a2+ab+b2)

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.

Let f(t)=t41t31. (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)41(1)31=111100

Note 2:

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

limt1(t31)=limt1(t3)limt11 (by limit law 2)=(1)31 (by limit law 9 and 7)=11=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 t1.

Simplify f(t) by using elementary algebra

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