Question

The value of $\displaystyle\lim _{x \rightarrow \pi / 2} \tan ^{2} x\left(\sqrt{2 \sin ^{2} x+3 \sin x+4}\right. -\sqrt{\left.\sin ^{2} x+6 \sin x+2\right)}$ is equal to

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

Answer 1

Answer: (C)

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