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

The correct option is C $\frac{1}{12}$

$\underset{x\to \pi /2}{lim}{\tan }^{2}x(\sqrt{2{\sin }^{2}x+3\sin x+4}-\sqrt{{\sin }^{2}x+6\sin x+2})$
Rationalizing,
$=\underset{x\to \pi /2}{lim}{\tan }^{2}x\frac{{\sin }^{2}x-3\sin x+2}{\sqrt{2{\sin }^{2}x+3\sin x+4}+\sqrt{{\sin }^{2}x+6\sin x+2}}$
$=\underset{x\to \pi /2}{lim}{\tan }^{2}x\frac{(2-\sin x)(1-\sin x)}{\sqrt{2{\sin }^{2}x+3\sin x+4}+\sqrt{{\sin }^{2}x+6\sin x+2}}$
$=\underset{x\to \pi /2}{lim}{\tan }^{2}x\frac{(2-\sin x){\cos }^{2}x}{(1+\sin x)[\sqrt{2{\sin }^{2}x+3\sin x+4}+\sqrt{{\sin }^{2}x+6\sin x+2}]}$
$=\underset{x\to \pi /2}{lim}\frac{{\sin }^{2}x(2-\sin x)}{(1+\sin x)(\sqrt{2{\sin }^{2}x+3\sin x+4}+\sqrt{{\sin }^{2}x+6\sin x+2})}=\frac{1}{12}$