Question

If $f(x)=x \tan ^{-1} x$, then $\displaystyle\lim _{x \rightarrow 1} \frac{f(x)-f(1)}{x-1}$ equals to

• A. $\frac{\pi+3}{4}$
• B. $\frac{\pi}{4}$
• C. $\frac{\pi+1}{4}$
• D. $\frac{\pi+2}{4}$

Answer 1

Answer: (D)

Given, $f(x)=x \tan ^{-1} x$ then
Then, $\displaystyle\lim _{x \rightarrow 1} \frac{f(x)-f(1)}{x-1}\left(\frac{0}{0}\right.$ form $)$
$=\displaystyle\lim _{x \rightarrow 1} \frac{f^{'}(x)-0}{1}\,\,\,$(using L'Hospital rule)
$=\displaystyle\lim _{x \rightarrow 1}\left(\frac{x}{1+x^{2}}+\tan ^{-1} x\right)$
$=\frac{1}{1+1^{2}}+\tan ^{-1}=\frac{1}{2}+\frac{\pi}{4}$
$=\frac{2+\pi}{4}$ or $\frac{\pi+2}{4}$