Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 2: Limits and Continuity - Section 2.2 - Limit of a Function and Limit Laws - Exercises 2.2 - Page 57: 34



Work Step by Step

We want to find $\lim\limits_{v \to 2}\dfrac{v^3-8}{v^4-16}$, but we can't use the quotient rule for limits because the limit of the denominator as $v$ approaches 2 is zero, and if we try to substitute $v=2$ directly, we get zero in both the numerator and the denominator. But note that $\dfrac{v^3-8}{v^4-16}=\dfrac{(v-2)(v^2+2v+4)}{(v^2-4)(v^2+4)}=\dfrac{(v-2)(v^2+2v+4)}{(v-2)(v+2)(v^2+4)}=\dfrac{v^2+2v+4}{(v+2)(v^2+4)}$ for all $v \neq 2.$ Thus $\lim\limits_{v \to 2}\dfrac{v^3-8}{v^4-16}=\lim\limits_{v \to 2}\dfrac{v^2+2v+4}{(v+2)(v^2+4)}=\dfrac{4+4+4}{(4)(8)}=\dfrac{12}{4(8)}=\dfrac{3}{8}.$
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