Question

If $f\left(9\right)=9$ and $f^{\prime }\left(9\right)=4$, then $\underset{x\to 9}{lim}\frac{\sqrt{f\left(x\right)}-3}{\sqrt{x}-3}$ is equal to

Answer 1

$\underset{x\to 9}{lim}\frac{\sqrt{f\left(x\right)}-3}{\sqrt{x}-3}=\underset{x\to 9}{lim}\frac{f\left(x\right)-9}{x-9}.\frac{\sqrt{x}+3}{\sqrt{f\left(x\right)}+3}$
$=f^{\prime }\left(9\right).\frac{3+3}{3+3}=f^{\prime }\left(9\right)=4$