Q61 / 160 · AP EAMCET 2014 · Mathematics
If $f(x)=x \tan ^{-1} x$, then $\lim _{x \rightarrow 1} \frac{f(x)-f(1)}{x-1}$ equals to
- A$\frac{\pi+3}{4}$
- B$\frac{\pi}{4}$
- C$\frac{\pi+1}{4}$
- Correct answerD$\frac{\pi+2}{4}$
Solution
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)
$$
\begin{aligned}
& =\lim _{x \rightarrow 1}\left(\frac{x}{1+x^2}+\tan ^{-1} x\right) \\
& =\frac{1}{1+1^2}+\tan ^{-1}=\frac{1}{2}+\frac{\pi}{4} \\
& =\frac{2+\pi}{4} \text { or } \frac{\pi+2}{4}
\end{aligned}
$$
Hence, the correct answer is option: 'D'.