Answer
$$\lim_{x\to9}\frac{\sin(\sqrt x-3)}{x-9}=\frac{1}{6}$$
Work Step by Step
$$A=\lim_{x\to9}\frac{\sin(\sqrt x-3)}{x-9}$$
$$A=\lim_{x\to9}\frac{\sin(\sqrt x-3)}{\sqrt x-3}\times\lim_{x\to9}\frac{\sqrt x-3}{x-9}$$
As $x\ne9$ then $\sqrt x-3\ne0$, and since $\lim_{x\to9}(\sqrt x-3)=\sqrt9-3=0$, we have $\lim_{x\to9}\frac{\sin(\sqrt x-3)}{\sqrt x-3}=1$.
$$A=1\times\lim_{x\to9}\frac{\sqrt x-3}{x-9}$$
$$A=\lim_{x\to9}\frac{\sqrt x-3}{(\sqrt x)^2-3^2}$$
$$A=\lim_{x\to9}\frac{\sqrt x-3}{(\sqrt x-3)(\sqrt x+3)}$$
$$A=\lim_{x\to9}\frac{1}{\sqrt x+3}$$
$$A=\frac{1}{\sqrt9+3}=\frac{1}{6}$$
