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If $f(x)=x \tan ^{-1} x$, then $\lim _{x \rightarrow 1} \frac{f(x)-f(1)}{x-1}$ equals to
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Verified Answer
The correct answer is:
$\frac{\pi+2}{4}$
Given, $f(x)=x \tan ^{-1} x$ then Then, $\lim _{x \rightarrow 1} \frac{f(x)-f(1)}{x-1}\left(\frac{0}{0}\right.$ form $)$
$$
=\lim _{x \rightarrow 1} \frac{f^{\prime}(x)-0}{1}
$$
(using L'Hospital rule)
$$
\begin{aligned}
& =\lim _{x \rightarrow 1}\left(\frac{x}{1+x^2}+\tan ^{-1} x\right) \\
& =\frac{1}{1+1^2}+\tan ^{-1}=\frac{1}{2}+\frac{\pi}{4} \\
& =\frac{2+\pi}{4} \text { or } \frac{\pi+2}{4}
\end{aligned}
$$
$$
=\lim _{x \rightarrow 1} \frac{f^{\prime}(x)-0}{1}
$$
(using L'Hospital rule)
$$
\begin{aligned}
& =\lim _{x \rightarrow 1}\left(\frac{x}{1+x^2}+\tan ^{-1} x\right) \\
& =\frac{1}{1+1^2}+\tan ^{-1}=\frac{1}{2}+\frac{\pi}{4} \\
& =\frac{2+\pi}{4} \text { or } \frac{\pi+2}{4}
\end{aligned}
$$
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