Mathematics - Limits Question with Solution | TestHub

We have:

$$\mathrm{LHL}=\underset{x\to \frac{\pi }{2}-}{\lim }\sqrt{\frac{\tan x-\sin \left({\tan }^{-1}(\tan x)\right)}{\tan x+{\cos }^2(\tan x)}}$$

$$=\underset{x\to \frac{\pi }{2}-}{\lim }\sqrt{\frac{\tan x-\sin x}{\tan x+{\cos }^2(\tan x)}}$$

$$=\underset{x\to \frac{\pi }{2}-}{\lim }\sqrt{\frac{1-\frac{\sin x}{\tan x}}{1+\frac{{\cos }^2(\tan x)}{\tan x}}}=\sqrt{\frac{1-0}{1+0}}=1$$

At $x\to \frac{\pi }{2}-$, $0<x<\frac{\pi }{2}$

${\tan }^{-1}(\tan x)=x$.

Further, as $x\to \frac{\pi }{2}-$, $\tan x\to \infty$, and ${\cos }^2(\tan x)$ is a real number between 0 and 1.

$$\mathrm{RHL}=\underset{x\to \frac{\pi }{2}+}{\lim }\sqrt{\frac{\tan x-\sin \left({\tan }^{-1}(\tan x)\right)}{\tan x+{\cos }^2(\tan x)}}$$

$$=\underset{x\to \frac{\pi }{2}+}{\lim }\sqrt{\frac{\tan x+\sin x}{\tan x+{\cos }^2(\tan x)}}$$

$$=\underset{x\to \frac{\pi }{2}+}{\lim }\sqrt{\frac{1+\frac{\sin x}{\tan x}}{1+\frac{{\cos }^2(\tan x)}{\tan x}}}=\sqrt{\frac{1+0}{1+0}}=1$$

(As $x\to \frac{\pi }{2}+$, $x>\frac{\pi }{2}\Rightarrow {\tan }^{-1}(\tan x)={\tan }^{-1}(\tan (x-\pi ))=x-\pi$.

$\sin \left({\tan }^{-1}(\tan x)\right)=\sin (x-\pi )=-\sin x$).

Further, as $x\to \frac{\pi }{2}+$, $\tan x\to -\infty$, and ${\cos }^2(\tan x)$ is a real number between 0 and 1.

$\mathrm{LHL}=\mathrm{RHL}=1$, the required limit is $1$.