Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 4 - Applications of the Derivative - 4.7 L'Hopital's Rule - 4.7 Exercises: 33

Answer

The solution is $$\lim_{v\to3}\frac{v-1-\sqrt{v^2-5}}{v-3}=-\frac{1}{2}.$$

Work Step by Step

To solve this limit follow the steps below $$\lim_{v\to3}\frac{v-1-\sqrt{v^2-5}}{v-3}=\lim_{v\to3}\frac{v-1-\sqrt{v^2-5}}{v-3}\cdot\frac{v-1+\sqrt{v^2-5}}{v-1+\sqrt{v^2-5}}=\lim_{v\to3}\frac{(v-1)^2-\sqrt{v^2-5}^2}{(v-3)(v-1+\sqrt{v^2-5})}=\lim_{v\to3}\frac{v^2+1-2v-v^2+5}{(v-3)(v-1+\sqrt{v^2-5})}=\lim_{v\to3}\frac{-2v+6}{(v-3)(v-1+\sqrt{v^2-5})}=\lim_{v\to3}\frac{-2(v-3)}{(v-3)(v-1+\sqrt{v^2-5})}=\lim_{v\to3}\frac{-2}{v-1+\sqrt{v^2-5}}=\frac{-2}{3-1+\sqrt{3^2-5}}=\frac{-2}{2+2}=-\frac{1}{2}.$$
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