Question

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

• A. 9
• B. 4
• C. 36
• D. None of these

Answer 1

Answer: (B)

$\underset{x\to 9}{\lim }\frac{\sqrt{f(x)}-3}{\sqrt{x}-3}$
$=\underset{x\to 9}{\lim }\frac{\sqrt{f(x)}-3}{\sqrt{x}-3}\cdot \frac{\sqrt{x}+3}{\sqrt{x}+3}\cdot \frac{\sqrt{f(x)}+3}{\sqrt{f(x)}+3}$
$=\underset{x\to 9}{\lim }\frac{f(x)-9}{x-9}\cdot \frac{\sqrt{x}+3}{\sqrt{f(x)}+3}$
$=\underset{x\to 9}{\lim }{f}^{\prime }(x)\frac{\sqrt{x}+3}{\sqrt{f(x)}+3}$
$=f^{\prime }(9)\frac{\sqrt{9}+3}{\sqrt{f(9)}+3}$
$=4\cdot \frac{6}{6}=4$