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

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Question

(B) We need to evaluate the limit: $L = \lim _{x ightarrow 0} \frac{x 	an 2x - 2x 	an x}{(1 - \cos 2x)^2}$.Using the identity $1 - \cos 2x = 2 \si

Vedclass · Accepted Answer

$\frac{1}{2}$

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(B) We need to evaluate the limit: $L = \lim _{x ightarrow 0} \frac{x 	an 2x - 2x 	an x}{(1 - \cos 2x)^2}$.Using the identity $1 - \cos 2x = 2 \sin^2 x$,the denominator becomes $(2 \sin^2 x)^2 = 4 \sin^4 x$.So,$L = \lim _{x ightarrow 0} \frac{x 	an 2x - 2x 	an x}{4 \sin^4 x}$.Using the Taylor series expansions for $	an 	heta \approx 	heta + \frac{	heta^3}{3}$ and $\sin 	heta \approx 	heta$:$	an 2x \approx 2x + \frac{(2x)^3}{3} = 2x + \frac{8x^3}{3}$.$	an x \approx x + \frac{x^3}{3}$.$\sin x \approx x$,so $\sin^4 x \approx x^4$.Substituting these into the expression:$L = \lim _{x ightarrow 0} \frac{x(2x + \frac{8x^3}{3}) - 2x(x + \frac{x^3}{3})}{4x^4}$.$L = \lim _{x ightarrow 0} \frac{2x^2 + \frac{8x^4}{3} - 2x^2 - \frac{2x^4}{3}}{4x^4}$.$L = \lim _{x ightarrow 0} \frac{\frac{6x^4}{3}}{4x^4} = \lim _{x ightarrow 0} \frac{2x^4}{4x^4} = \frac{2}{4} = \frac{1}{2}$.

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

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