Question
The value of $$\mathop {\lim }\limits_{x \to \frac{\pi }{2}} {\tan ^2}x\left( {\sqrt {2\,{{\sin }^2}x + 3\sin \,x + 4} - \sqrt {{{\sin }^2}x + 6\sin \,x + 2} } \right)$$ is equal to :
A.
$$\frac{1}{{10}}$$
B.
$$\frac{1}{{11}}$$
C.
$$\frac{1}{{12}}$$
D.
$$\frac{1}{8}$$
Answer :
$$\frac{1}{{12}}$$
Solution :
$$\eqalign{
& \mathop {\lim }\limits_{x \to \frac{\pi }{2}} {\tan ^2}x\left( {\sqrt {2\,{{\sin }^2}x + 3\sin \,x + 4} - \sqrt {{{\sin }^2}x + 6\sin \,x + 2} } \right) \cr
& = \mathop {\lim }\limits_{x \to \frac{\pi }{2}} {\tan ^2}x\frac{{\left( {2\,{{\sin }^2}x + 3\sin \,x + 4 - \,{{\sin }^2}x - 6\sin \,x - 2} \right)}}{{\sqrt {2\,{{\sin }^2}x + 3\sin \,x + 4} + \sqrt {{{\sin }^2}x + 6\sin \,x + 2} }} \cr
& = \mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{{{\tan }^2}x\left( {{{\sin }^2}x - 3\sin \,x + 2} \right)}}{{\sqrt {2\,{{\sin }^2}x + 3\sin \,x + 4} + \sqrt {{{\sin }^2}x + 6\sin \,x + 2} }} \cr
& = \mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{{{\sin }^2}x\left( {\sin \,x - 1} \right)\left( {\sin \,x - 2} \right)}}{{\left( {1 - {{\sin }^2}x} \right)\left( {\sqrt {2\,{{\sin }^2}x + 3\sin \,x + 4} + \sqrt {{{\sin }^2}x + 6\sin \,x + 2} } \right)}} \cr
& = \mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{ - {{\sin }^2}x\left( {\sin \,x - 2} \right)}}{{\left( {1 - \sin x} \right)\left( {\sqrt {2\,{{\sin }^2}x + 3\sin \,x + 4} + \sqrt {{{\sin }^2}x + 6\sin \,x + 2} } \right)}} \cr
& = \frac{1}{{2\left( {\sqrt 9 + \sqrt 9 } \right)}} \cr
& = \frac{1}{{12}} \cr} $$