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$\lim\limits_{x\to 0}\large\frac{x\tan 2x-2x\tan x}{(1-\cos 2x)^2}$ is

$(a)\;2\qquad(b)\;-2\qquad(c)\;1/2\qquad(d)\;-1/2$

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$\lim\limits_{x\to 0}\large\frac{x\tan 2x-2x\tan x}{4\sin^4x}$
$\Rightarrow \lim\limits_{x\to 0}\large\frac{x}{4\sin^4x}\bigg[\large\frac{2\tan x}{1-\tan^2x}$$-2\tan x\bigg]$
$\Rightarrow \lim\limits_{x\to 0}\large\frac{2x\tan x}{4\sin^4x}\bigg[\large\frac{1-1+\tan^2x}{1-\tan^2x}\bigg]$
$\Rightarrow \lim\limits_{x\to 0}\large\frac{2x\tan^3x}{4\sin^4x(1-\tan^2x)}$
$\Rightarrow\large\frac{1}{2}$$ \lim\limits_{x\to 0}\large\frac{x}{\sin x}.\frac{1}{\cos^3x}.\frac{1}{1-\tan^2x}$
$\Rightarrow \large\frac{1}{2}$$.1.\large\frac{1}{1^3}.\frac{1}{1-0}$
$\Rightarrow \large\frac{1}{2}$
Hence (c) is the correct answer.
answered Jan 3, 2014 by sreemathi.v
 

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