If f(9)=9 and f^{prime}(9)=4,then lim _{x rig | vedclass

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(B) Given the limit is in the $\frac{0}{0}$ form,we apply $L'H\hat{o}pital$'s rule:$\lim _{x \rightarrow 9} \frac{\sqrt{f(x)}-3}{\sqrt{x}-3} = \lim _{

Vedclass · Accepted Answer

$4$

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(B) Given the limit is in the $\frac{0}{0}$ form,we apply $L'H\hat{o}pital$'s rule:$\lim _{x \rightarrow 9} \frac{\sqrt{f(x)}-3}{\sqrt{x}-3} = \lim _{x \rightarrow 9} \frac{\frac{d}{dx}(\sqrt{f(x)}-3)}{\frac{d}{dx}(\sqrt{x}-3)}$$= \lim _{x \rightarrow 9} \frac{\frac{f^{\prime}(x)}{2\sqrt{f(x)}}}{\frac{1}{2\sqrt{x}}}$$= \lim _{x \rightarrow 9} \frac{f^{\prime}(x) \cdot \sqrt{x}}{\sqrt{f(x)}}$Substituting the values $f(9)=9$ and $f^{\prime}(9)=4$:$= \frac{f^{\prime}(9) \cdot \sqrt{9}}{\sqrt{f(9)}} = \frac{4 \cdot 3}{\sqrt{9}} = \frac{12}{3} = 4$.

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

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