The value of $\lim _{x ightarrow \frac{\pi}{2}} \sqrt{\frac{ an x-\sin \left\{ an ^{-1}( an x)\

Question

The value of $\lim _{x ightarrow \frac{\pi}{2}} \sqrt{\frac{	an x-\sin \left\{	an ^{-1}(	an x)ight\}}{	an x+\cos ^2(	an x)}}$ is $\_\_\_\_$

Professor PC Thomas Classes · Answer

We have, $\lim_{xightarrow \frac{\pi}{2}} \sqrt{\frac{	an x - \sin\{	an^{-1}(	an x)\}}{	an x + \cos^2(	an x)}}$

Now, RHL at $x = \frac{\pi}{2}$,

$= \lim_{xightarrow \frac{\pi^+}{2}} \sqrt{\frac{	an x - \sin\{	an^{-1}(	an x)\}}{	an x + \cos^2(	an x)}} = \lim_{xightarrow \frac{\pi^+}{2}} \sqrt{\frac{	an x - \sin(x - \pi)}{	an x + \cos^2(	an x)}} = 1$

$\left[ \because 	an^{-1}(	an x) = x - \pi, 	ext{ when } x > \frac{\pi}{2} ight]$

Again, LHL at $x = \frac{\pi}{2}$,

$= \lim_{xightarrow \frac{\pi^-}{2}} \sqrt{\frac{	an x - \sin\{	an^{-1}(	an x)\}}{	an x + \cos^2(	an x)}} = \lim_{xightarrow \frac{\pi^-}{2}} \sqrt{\frac{	an x - \sin(x)}{	an x + \cos^2(	an x)}}$

$\left[ \because 	an^{-1}(	an x) = x, 	ext{ when } x < \frac{\pi}{2} ight]$

$= \lim_{xightarrow \frac{\pi^-}{2}} \sqrt{\frac{1 - \frac{\sin x}{	an x}}{1 + \frac{\cos^2(	an x)}{	an x}}} = \sqrt{\frac{1-0}{1+0}} = 1 = \lim_{xightarrow \frac{\pi}{2}} \sqrt{\frac{	an x - \sin\{	an^{-1}(	an x)\}}{	an x + \cos^2(	an x)}} = 1.$