Question

The value of $\underset{x\to \pi /2}{\lim }{\tan }^{2}x(\sqrt{2{\sin }^{2}x+3\sin x+4}-\sqrt{{\sin }^{2}x+6\sin x+2)}$ is equal to

• A. $\frac{1}{10}$
• B. $\frac{1}{11}$
• C. $\frac{1}{12}$
• D. $\frac{1}{8}$

Answer 1

Answer: (C)

$\underset{x\to \pi /2}{\lim }{\tan }^{2}x\left(\sqrt{2{\sin }^{2}x+3\sin x+4}-\sqrt{{\sin }^{2}x+6\sin x+2}\right)$
$=\underset{x\to \pi /2}{\lim }{\tan }^{2}x\frac{\left(2{\sin }^{2}x+3\sin x+4-{\sin }^{2}x-6\sin x-2\right)}{\sqrt{2{\sin }^{2}x+3\sin x+4}+\sqrt{{\sin }^{2}x+6\sin x+2}}$
$=\underset{x\to \pi /2}{\lim }\frac{{\tan }^{2}x\left({\sin }^{2}x-3\sin x+2\right)}{\sqrt{2{\sin }^{2}x+3\sin x+4}+\sqrt{{\sin }^{2}x+6\sin x+2}}$
$=\underset{x\to \pi /2}{\lim }\frac{{\sin }^{2}x(\sin x-1)(\sin x-2)}{\left(1-{\sin }^{2}x\right)\left(\sqrt{2{\sin }^{2}x+3\sin x+4}+\sqrt{{\sin }^{2}x+6\sin x+2}\right)}$
$=\underset{x\to \pi /2}{\lim }\frac{-{\sin }^{2}x(\sin x-2)}{(1+\sin x)\left(\sqrt{2{\sin }^{2}x+3\sin x+4}+\sqrt{{\sin }^{2}x+6\sin x+2}\right)}$
$=\frac{1}{2(\sqrt{9}+\sqrt{9})}=\frac{1}{12}$