A) 2
B) ?2
C) \[\frac{1}{2}\]
D) \[-\frac{1}{2}\]
Correct Answer: C
Solution :
\[\underset{x\to 0}{\mathop{\lim }}\,\,\frac{x\tan 2x-2x\tan x}{{{(1-\cos \,\,2x)}^{2}}}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\,\frac{x(\tan \,\,2x-2\tan x)}{{{(2\,{{\sin }^{2}}x)}^{2}}}=\underset{x\to 0}{\mathop{\lim }}\,\,\,\frac{1}{4}\,\frac{x\,(\tan 2x-2\tan x)}{{{\sin }^{4}}x}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{1}{4}\frac{x\left\{ \left( 2x+\frac{1}{3}{{(2x)}^{3}}+\frac{2}{15}\,{{(2x)}^{5}}+... \right)-2\left( x+\frac{{{x}^{3}}}{3}+\frac{2}{15}{{x}^{5}}+... \right) \right\}}{{{x}^{4}}\,{{\left( 1-\frac{{{x}^{2}}}{3\,\,!}+\frac{{{x}^{4}}}{5\,\,!}+.... \right)}^{4}}}\] \[=\frac{1}{4}\,.\,\left( \frac{8}{3}-\frac{2}{3} \right)=\frac{2}{4}=\frac{1}{2}\].
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