Question

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

• A. $4$
• B. $\frac{1}{4}$
• C. $\frac{1}{2}$
• D. $-\frac{1}{2}$

Answer 1

The correct option is A $4$

$f\left(9\right)=9,f^{\prime }\left(9\right)=4$

$\underset{x\to 9}{lim}\frac{\sqrt{f\left(x\right)}-3}{\sqrt{x}-3}=\frac{0}{0}$

$\underset{x\to 9}{lim}\frac{\frac{f^{\prime }\left(x\right)}{2\sqrt{f\left(x\right)}}-0}{\frac{1}{2\sqrt{x}}-0}=\frac{\frac{f^{\prime }\left(9\right)}{\sqrt{f\left(9\right)}}}{\frac{1}{\sqrt{9}}}=f^{\prime }\left(9\right)$

$=4$