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

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

Aicharya · Accepted Answer

Given, $f(x)=x 	an ^{-1} x$ then Then, $\lim _{x ightarrow 1} \frac{f(x)-f(1)}{x-1}\left(\frac{0}{0}ight.$ form $)$ $ =\lim _{x ightarrow 1} \frac{f^{\prime}(x)-0}{1} $ (using L'Hospital rule) $ \begin{aligned} & =\lim _{x ightarrow 1}\left(\frac{x}{1+x^2}+	an ^{-1} xight) \ & =\frac{1}{1+1^2}+	an ^{-1}=\frac{1}{2}+\frac{\pi}{4} \ & =\frac{2+\pi}{4} 	ext { or } \frac{\pi+2}{4} \end{aligned} $

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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