If $f(9)=9, f^{\prime}(9)=4$ then $\underset{x \rightarrow 9}{\operatorname{Lt}} \frac{\sqrt{f(x)}-3

Question

If $f(9)=9, f^{\prime}(9)=4$ then $\underset{x \rightarrow 9}{\operatorname{Lt}} \frac{\sqrt{f(x)}-3}{\sqrt{x}-3}$ equals.........

Professor PC Thomas Classes · Answer

$\lim _{x ightarrow 9} \frac{\sqrt{f(x)}-3}{\sqrt{x}-3}=\lim _{x ightarrow 9} \frac{f(x)-9}{\sqrt{f(x)}+3} \cdot \frac{\sqrt{x}+3}{x-9}$\ $=\lim _{x ightarrow 9} \frac{f(x)-9}{x-9} \cdot \lim _{x ightarrow 9} \frac{\sqrt{x}+3}{\sqrt{f(x)}+3}$\ $=f^{\prime}(9)\left(\frac{\sqrt{9}+3}{\sqrt{f(9)}+3}ight)=4 \cdot\left(\frac{3+3}{3+3}ight)=4$\