A) \[\frac{1}{10}\]
B) \[\frac{1}{11}\]
C) \[\frac{1}{12}\]
D) \[\frac{1}{8}\]
Correct Answer: C
Solution :
[c] \[\underset{x\to \pi /2}{\mathop{\lim }}\,{{\tan }^{2}}x(\sqrt{2{{\sin }^{2}}x+3\sin x+4}-\sqrt{{{\sin }^{2}}x+6\sin x+2})\] \[=\underset{x\to \pi /2}{\mathop{\lim }}\,{{\tan }^{2}}x\frac{(2si{{n}^{2}}x+3sinx+4-si{{n}^{2}}x-6\sin x-2)}{\sqrt{2{{\sin }^{2}}x+3\sin x+4}+\sqrt{{{\sin }^{2}}x+6\sin x+2}}\] \[=\underset{x\to \pi /2}{\mathop{\lim }}\,\frac{{{\tan }^{2}}x(si{{n}^{2}}x-3sinx+2)}{\sqrt{2{{\sin }^{2}}x+3\sin x+4}+\sqrt{{{\sin }^{2}}x+6\sin x+2}}\] \[=\,\underset{x\to \pi /2}{\mathop{\lim }}\,\frac{{{\sin }^{2}}x(sin\,x-1)(sin\,x-2)}{(1-si{{n}^{2}}x)(\sqrt{2{{\sin }^{2}}x+3\sin x+4}+\sqrt{{{\sin }^{2}}x+6\sin x+2})}\] \[\underset{x\to \pi /2}{\mathop{\lim }}\,\frac{-{{\sin }^{2}}x(sinx-2)}{(1+sinx)(\sqrt{2{{\sin }^{2}}x+3\sin x+4}+\sqrt{{{\sin }^{2}}x+6\sin x+2})}\]\[=\frac{1}{2(\sqrt{9}+\sqrt{9})}=\frac{1}{12}.\]
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