Question

$\underset{x\to \frac{\pi }{2}}{\lim }\left({\tan }^{2}x\left({\left(2{\sin }^{2}x+3\sin x+4\right)}^{\frac{1}{2}}-{\left({\sin }^{2}x+6\sin x+2\right)}^{\frac{1}{2}}\right)\right)$ is equal to

• A. $\frac{1}{12}$
• B. $-\frac{1}{18}$
• C. $-\frac{1}{12}$
• D. $-\frac{1}{6}$

Answer 1

Answer: (A)

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