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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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Verified Answer
The correct answer is:
16
We have, $f(4)=4$ and $f^{\prime}(4)=16$
Now, consider $\lim _{x \rightarrow 4} \frac{\sqrt{f(x)}-2}{\sqrt{x}-2} \quad\left[\frac{0}{0}\right.$ form $]$
$$
\begin{aligned}
& =\lim _{x \rightarrow 4} \frac{\frac{1}{2 \sqrt{f(x)}} f^{\prime}(x)}{\frac{1}{2 \sqrt{x}}} \quad \text { [by using } L \text { Hospi } \\
& =\lim _{x \rightarrow 4} \frac{f^{\prime}(x) \cdot \sqrt{x}}{\sqrt{f(x)}}=\frac{f^{\prime}(4) \cdot 2}{\sqrt{f(4)}}=\frac{16 \cdot 2}{\sqrt{4}}=\frac{32}{2}=16
\end{aligned}
$$
[by using $L$ Hospital rule]
Now, consider $\lim _{x \rightarrow 4} \frac{\sqrt{f(x)}-2}{\sqrt{x}-2} \quad\left[\frac{0}{0}\right.$ form $]$
$$
\begin{aligned}
& =\lim _{x \rightarrow 4} \frac{\frac{1}{2 \sqrt{f(x)}} f^{\prime}(x)}{\frac{1}{2 \sqrt{x}}} \quad \text { [by using } L \text { Hospi } \\
& =\lim _{x \rightarrow 4} \frac{f^{\prime}(x) \cdot \sqrt{x}}{\sqrt{f(x)}}=\frac{f^{\prime}(4) \cdot 2}{\sqrt{f(4)}}=\frac{16 \cdot 2}{\sqrt{4}}=\frac{32}{2}=16
\end{aligned}
$$
[by using $L$ Hospital rule]
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