mathop {lim }limits_{x to 0} frac{{{{left( {1 | vedclass

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(C) Let $L = \mathop {\lim }\limits_{x 	o 0} \frac{{{{\left( {1 - \cos 2x} ight)}^2}}}{{2x	an x - x	an 2x}}$.Using the identity $1 - \cos 2x = 2\

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$-2$

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(C) Let $L = \mathop {\lim }\limits_{x 	o 0} \frac{{{{\left( {1 - \cos 2x} ight)}^2}}}{{2x	an x - x	an 2x}}$.Using the identity $1 - \cos 2x = 2\sin^2 x$,we have:$L = \mathop {\lim }\limits_{x 	o 0} \frac{{{{\left( {2\sin^2 x} ight)}^2}}}{{2x	an x - x	an 2x}} = \mathop {\lim }\limits_{x 	o 0} \frac{{4\sin^4 x}}{{2x	an x - x	an 2x}}$.Using the Taylor series expansions $\sin x = x - \frac{x^3}{6} + O(x^5)$ and $	an x = x + \frac{x^3}{3} + O(x^5)$:$L = \mathop {\lim }\limits_{x 	o 0} \frac{{4(x - \frac{x^3}{6})^4}}{{2x(x + \frac{x^3}{3}) - x(2x + \frac{(2x)^3}{3})}} = \mathop {\lim }\limits_{x 	o 0} \frac{{4x^4}}{{2x^2 + \frac{2x^4}{3} - 2x^2 - \frac{8x^4}{3}}} = \mathop {\lim }\limits_{x 	o 0} \frac{{4x^4}}{-\frac{6x^4}{3}} = \mathop {\lim }\limits_{x 	o 0} \frac{{4x^4}}{-2x^4} = -2$.

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

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