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arrow_back If $\mathrm{f}(9)=9, \mathrm{f}^{\prime}(9)=4$, then $\lim _{x \rightarrow 9} \dfrac{\sqrt{f(x)}-3}{\sqrt{x}-3}$

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If $\mathrm{f}(9)=9, \mathrm{f}^{\prime}(9)=4$, then $\lim _{x \rightarrow 9} \dfrac{\sqrt{f(x)}-3}{\sqrt{x}-3}$

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Best answer
Applying L - Hospitals rule,

$\lim _{x \rightarrow 9} \frac{\dfrac{1}{2 \sqrt{f(x)}} \cdot f^{\prime}(x)}{\dfrac{1}{2 \sqrt{2}}}$

$=\dfrac{\dfrac{f^{\prime}(9)}{\sqrt{f(9)}}}{\dfrac{1}{\sqrt{9}}}$

$=\dfrac{\dfrac{4}{3}}{\dfrac{1}{3}}$

$=4$
by Platinum
(119,128 points)

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