mathop {lim }limits_{x to 0} frac{{xtan 2x - | vedclass

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Question

(C) Let $L = \mathop {\lim }\limits_{x 	o 0} \frac{{x	an 2x - 2x	an x}}{{{{(1 - \cos 2x)}^2}}}$Using the identity $1 - \cos 2x = 2\sin^2 x$,we get:

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$\frac{1}{2}$

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(C) Let $L = \mathop {\lim }\limits_{x 	o 0} \frac{{x	an 2x - 2x	an x}}{{{{(1 - \cos 2x)}^2}}}$Using the identity $1 - \cos 2x = 2\sin^2 x$,we get:$L = \mathop {\lim }\limits_{x 	o 0} \frac{{x(	an 2x - 2	an x)}}{{{{(2\sin^2 x)}^2}}} = \mathop {\lim }\limits_{x 	o 0} \frac{{x(	an 2x - 2	an x)}}{{4\sin^4 x}}$Using the Taylor series expansion $	an 	heta = 	heta + \frac{	heta^3}{3} + \dots$ and $\sin x \approx x$ as $x 	o 0$:$L = \mathop {\lim }\limits_{x 	o 0} \frac{{x \left[ (2x + \frac{(2x)^3}{3} + \dots) - 2(x + \frac{x^3}{3} + \dots) ight]}}{{4x^4}}$$L = \mathop {\lim }\limits_{x 	o 0} \frac{{x \left[ 2x + \frac{8x^3}{3} - 2x - \frac{2x^3}{3} ight]}}{{4x^4}}$$L = \mathop {\lim }\limits_{x 	o 0} \frac{{x \cdot (2x^3)}}{{4x^4}} = \mathop {\lim }\limits_{x 	o 0} \frac{{2x^4}}{{4x^4}} = \frac{2}{4} = \frac{1}{2}$.

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

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