Mathematics - Limits Question with Solution | TestHub

We have

$$\begin{array}{rl}& LHL=\underset{x\to \frac{\pi }{2}-}{\lim }\sqrt{\frac{\tan x-\sin {\tan }^{-1}(\tan x)}{\tan x+{\cos }^2(\tan x)}}\\ & =\underset{x\to \frac{\pi }{2}-}{\lim }\sqrt{\frac{\tan x-\sin x}{\tan x+{\cos }^2(\tan x)}}\\ & =\underset{x\to \frac{\pi }{2}-}{\lim }\sqrt{\frac{1-\frac{\sin x}{\tan x}}{1+\frac{{\cos }^2(\tan x)}{\tan x}}}=\sqrt{\frac{1-0}{1+0}}=1\end{array}$$

At $x\to \frac{\pi }{2}-,o<x<\frac{\pi }{2}\therefore {\tan }^{-1}(\tan x)=x$

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

$$\begin{array}{rl}\text{ RHL }=& \underset{x\to \frac{\pi }{2}+}{\lim }\sqrt{\frac{\tan x-\sin {\tan }^{-1}(\tan x)}{\tan x+{\cos }^2(\tan x)}}\\ =& \underset{x\to \frac{\pi }{2}+}{\lim }\sqrt{\frac{\tan x+\sin x}{\tan x+{\cos }^{2}x(\tan x)}}\\ =& \underset{x\to \frac{\pi }{2}+}{\lim }\sqrt{\frac{1+\frac{\sin x}{\tan x}}{1+\frac{{\cos }^2(\tan x)}{\tan x}}}=\sqrt{\frac{1+0}{1-0}}=1\\ & (\text{ As }x\to \frac{\pi }{2}+,x>\frac{\pi }{2}\Rightarrow {\tan }^{-1}\tan x\\ =& {\tan }^{-1}{\tan }^{\tan }(x-\pi )=x-\pi \\ & \therefore {\sin }^{-1}(\tan x)=\sin (x-\pi )=-\sin x\end{array}$$

Further as $x\to \frac{\pi }{2}+;\tan x\to -\infty$ and ${\cos }^2(\tan x)$ is a real number between 0 and 1) $\mathrm{LHL}=\mathrm{RHL}=1\therefore$ required limit $=1$