mathop {lim }limits_{x to 0} frac{{{{tan }^{ | vedclass

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(D) We are given the limit: $\mathop {\lim }\limits_{x 	o 0} \frac{{{{	an }^{ - 1}}x}}{x}$.Substituting $x = 0$,we get the indeterminate form $\frac

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(D) We are given the limit: $\mathop {\lim }\limits_{x 	o 0} \frac{{{{	an }^{ - 1}}x}}{x}$.Substituting $x = 0$,we get the indeterminate form $\frac{0}{0}$.Using $L$'$H$ôpital's rule,we differentiate the numerator and the denominator with respect to $x$:$\frac{d}{dx}(	an^{-1}x) = \frac{1}{1+x^2}$ and $\frac{d}{dx}(x) = 1$.Thus,$\mathop {\lim }\limits_{x 	o 0} \frac{\frac{1}{1+x^2}}{1} = \mathop {\lim }\limits_{x 	o 0} \frac{1}{1+x^2}$.Evaluating the limit as $x 	o 0$,we get $\frac{1}{1+0^2} = 1$.

$\mathop {\lim }\limits_{x \to 0} \frac{{{{\tan }^{ - 1}}x}}{x}$ is

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