Question

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

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

Answer 1

Answer: (C)

$\frac{1}{12}$

Let $L=\underset{x\to \frac{\pi }{2}}{lim}{\tan }^{2}x(\sqrt{2{\sin }^{2}x+3\sin x+4}-\sqrt{{\sin }^{2}x+6\sin x+2})$

Rationalizing, we get

$L=\underset{x\to \frac{\pi }{2}}{lim}\frac{{\sin }^{2}x-3\sin x+2}{{\cot }^{2}x(\sqrt{2{\sin }^{2}x+3\sin x+4}+\sqrt{{\sin }^{2}x+6\sin x+2})}$

Substituting limit in denominator ;

$L=\underset{x\to \frac{\pi }{2}}{lim}\frac{1}{6}\frac{{\sin }^{2}x-3\sin x+2}{{\cot }^{2}x}$

Limit is in the form 0/0 ;applying L'Hopitals rule we get ;

$L=\underset{x\to \frac{\pi }{2}}{lim}\frac{1}{6}\frac{2\sin x\cos x-3\cos x}{-2\cot x{\text{cosec}}^{2}x}$

Cancelling $\cos x$ and rewriting the limit ;

$L=\underset{x\to \frac{\pi }{2}}{lim}\frac{1}{6}\frac{2\sin x-3}{-2}\left({\sin }^{3}x\right)$

Substituting $x=\frac{\pi }{2}$ in limit, we get ;

$L=\frac{1}{12}$