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If $\mathrm{f}(9)=9$ and $\mathrm{f}^{\prime}(9)=4$, then $\lim _{x \rightarrow 9} \frac{\sqrt{f(x)}-3}{\sqrt{x}-3}=$
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$4$
$\begin{aligned} & \lim _{x \rightarrow 9} \frac{\sqrt{f(x)}-3}{\sqrt{x}-3} \\ = & \lim _{x \rightarrow 9} \frac{\frac{f^{\prime}(x)}{2 \sqrt{f(x)}}}{\frac{1}{2 \sqrt{x}}} \quad\left(\frac{0}{0} \text { form }\right) \\ = & \lim _{x \rightarrow 9} \frac{f^{\prime}(x)}{\sqrt{f(x)}} \times \sqrt{x}\end{aligned}$
$=\frac{f^{\prime}(9)}{\sqrt{f(9)}} \times \sqrt{9}=\frac{4}{3} \times 4=4$.
$=\frac{f^{\prime}(9)}{\sqrt{f(9)}} \times \sqrt{9}=\frac{4}{3} \times 4=4$.
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