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 - Additional and Advanced Exercises - Page 104: 30



Work Step by Step

\begin{align*} \lim _{x \rightarrow 9} \frac{\sin (\sqrt{x}-3)}{x-9}&=\lim _{x \rightarrow 9} \frac{\sin (\sqrt{x}-3)}{\sqrt{x}-3} \cdot \frac{1}{\sqrt{x}+3}\\ &=\lim _{x \rightarrow 9} \frac{\sin (\sqrt{x}-3)}{\sqrt{x}-3} \cdot \lim _{x \rightarrow 9} \frac{1}{\sqrt{x}+3}\\ &=1 \cdot \frac{1}{6}=\frac{1}{6}\\ &=\frac{1}{6} \end{align*}
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