lim _{x rightarrow 0} frac{x tan 2x - 2x tan | vedclass

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(C) We evaluate the limit: $\lim _{x ightarrow 0} \frac{x 	an 2x - 2x 	an x}{(1 - \cos 2x)^2}$.Using the identity $1 - \cos 2x = 2 \sin^2 x$,the d

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$\frac{1}{2}$

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(C) We evaluate the limit: $\lim _{x ightarrow 0} \frac{x 	an 2x - 2x 	an x}{(1 - \cos 2x)^2}$.Using the identity $1 - \cos 2x = 2 \sin^2 x$,the denominator becomes $(2 \sin^2 x)^2 = 4 \sin^4 x$.So,the expression is $\lim _{x ightarrow 0} \frac{x(	an 2x - 2 	an x)}{4 \sin^4 x}$.Using Taylor series expansions: $	an 	heta = 	heta + \frac{	heta^3}{3} + \dots$ and $\sin x \approx x$.$	an 2x = 2x + \frac{(2x)^3}{3} + O(x^5) = 2x + \frac{8x^3}{3} + O(x^5)$.$2 	an x = 2(x + \frac{x^3}{3} + O(x^5)) = 2x + \frac{2x^3}{3} + O(x^5)$.Numerator: $x[(2x + \frac{8x^3}{3}) - (2x + \frac{2x^3}{3})] = x[\frac{6x^3}{3}] = 2x^4$.Denominator: $4 \sin^4 x \approx 4x^4$.Limit: $\lim _{x ightarrow 0} \frac{2x^4}{4x^4} = \frac{2}{4} = \frac{1}{2}$.

$\lim _{x \rightarrow 0} \frac{x \tan 2x - 2x \tan x}{(1 - \cos 2x)^2}$ is

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