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Let $f(x)$ be a differentiable function and $f^{\prime}(4)=5 .$ Then, $\lim _{x \rightarrow 2} \frac{f(4) - f\left(x^{2}\right)}{x-2}$ equals
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The correct answer is:
-20
Given, $f^{\prime}(4)=5$
$$
\begin{array}{ll}
\text { Now, } \lim _{x \rightarrow 2} \frac{f(4)-f\left(x^{2}\right)}{x-2} & {\left[\frac{0}{0} \text { form }\right]} \\
=\lim _{x \rightarrow 2} \frac{0-f^{\prime}\left(x^{2}\right) \cdot 2 x}{1} \\
=\frac{-f^{\prime}(4) \cdot 2 \times 2}{1} \\
=-(5) \times 4=-20
\end{array}
$$
$$
\begin{array}{ll}
\text { Now, } \lim _{x \rightarrow 2} \frac{f(4)-f\left(x^{2}\right)}{x-2} & {\left[\frac{0}{0} \text { form }\right]} \\
=\lim _{x \rightarrow 2} \frac{0-f^{\prime}\left(x^{2}\right) \cdot 2 x}{1} \\
=\frac{-f^{\prime}(4) \cdot 2 \times 2}{1} \\
=-(5) \times 4=-20
\end{array}
$$
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