Chapter 2: Limits and Continuity - Additional and Advanced Exercises - Page 104: 30

$$\frac{1}{6}$$

\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*}