Mathematics - Limits Question with Solution | TestHub

Given, $\underset{x\to \frac{\pi }{2}}{\lim }{\tan }^{2}x\left[\sqrt{2{\sin }^{2}x+3\sin x+4}-\sqrt{{\sin }^{2}x+6\sin x+2}\right]$

$=\underset{x\to \frac{\pi }{2}}{\lim }{\tan }^{2}x\frac{\left[{\left(\sqrt{2{\sin }^{2}x+3\sin x+4}\right)}^2-{\left(\sqrt{{\sin }^{2}x+6\sin x+2}\right)}^2\right]}{\sqrt{2{\sin }^{2}x+3\sin x+4}+\sqrt{{\sin }^{2}x+6\sin x+2}}$

$=\underset{x\to \frac{\pi }{2}}{\lim }\frac{{\tan }^{2}x\left[{\sin }^{2}x-3\sin x+2\right]}{\sqrt{9}+\sqrt{9}}$

$=\underset{x\to \frac{\pi }{2}}{\lim }\frac{{\tan }^{2}x\left(\sin x-1\right)\left(\sin x-2\right)}{6}$

$=\frac{1}{6}\underset{x\to \frac{\pi }{2}}{\lim }{\tan }^{2}x\left(1-\sin x\right)$

$=\frac{1}{6}\underset{x\to \frac{\pi }{2}}{\lim }\frac{{\sin }^{2}x\left(1-\sin x\right)}{\left(1-\sin x\right)\left(1+\sin x\right)}=\frac{1}{12}$