If f is a real function such that f(4)=4 and | vedclass

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(A) Given $f(4)=4$ and $f^{\prime}(4)=16$. Consider the limit $L = \lim _{x \rightarrow 4} \frac{\sqrt{f(x)}-2}{\sqrt{x}-2}$. This is a $\left[\frac{0

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$16$

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(A) Given $f(4)=4$ and $f^{\prime}(4)=16$. Consider the limit $L = \lim _{x ightarrow 4} \frac{\sqrt{f(x)}-2}{\sqrt{x}-2}$. This is a $\left[\frac{0}{0}ight]$ indeterminate form. Applying $L'	ext{Hospital's rule}$,we differentiate the numerator and denominator with respect to $x$: $L = \lim _{x ightarrow 4} \frac{\frac{d}{dx}(\sqrt{f(x)}-2)}{\frac{d}{dx}(\sqrt{x}-2)} = \lim _{x ightarrow 4} \frac{\frac{1}{2\sqrt{f(x)}} \cdot f^{\prime}(x)}{\frac{1}{2\sqrt{x}}}$. Simplifying the expression: $L = \lim _{x ightarrow 4} \frac{f^{\prime}(x) \cdot \sqrt{x}}{\sqrt{f(x)}} = \frac{f^{\prime}(4) \cdot \sqrt{4}}{\sqrt{f(4)}}$. Substituting the given values: $L = \frac{16 \cdot 2}{\sqrt{4}} = \frac{32}{2} = 16$.

If $f$ is a real function such that $f(4)=4$ and $f^{\prime}(4)=16$,then $\lim _{x \rightarrow 4} \frac{\sqrt{f(x)}-2}{\sqrt{x}-2} =$

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