Question

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

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

Given that f(9) = 9, f’(9) = 4

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

On rationalisation we get -

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

$=\underset{x\to 9}{lim}\frac{f\left(x\right)-f\left(9\right)}{x-9}$ . $\frac{6}{6}$

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

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

$=4$