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# If $f(9)=9,f'(9)=4$ then $\lim\limits_{x\to 9}\large\frac{\sqrt{f(x)}-3}{\sqrt x-3}$ is

$(a)\;4\qquad(b)\;0\qquad(c)\;2\qquad(d)\;9$

$\lim\limits_{x\to 9}\large\frac{\sqrt{f(x)}-3}{\sqrt x-3}\qquad\big(\large\frac{0}{0}\big)$
By L Hospital rule
$\Rightarrow \lim\limits_{x\to 9}\large\frac{\Large\frac{f'(x)}{2\sqrt{f(x)}}}{1/2\sqrt x}$
$\Rightarrow \lim\limits_{x\to 9}\large\frac{\sqrt xf'(x)}{\sqrt{f(x)}}$
$\Rightarrow \large\frac{3\times 4}{3}$
$\Rightarrow 4$
Hence (a) is the correct answer.