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

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(D) We use the Taylor series expansion for $	an x$ and $\cos 2x$ near $x=0$: $	an x = x + \frac{x^3}{3} + \frac{2x^5}{15} + \dots$ $	an 2x = 2x + \

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$\frac{1}{4}$

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(D) We use the Taylor series expansion for $	an x$ and $\cos 2x$ near $x=0$: $	an x = x + \frac{x^3}{3} + \frac{2x^5}{15} + \dots$ $	an 2x = 2x + \frac{(2x)^3}{3} + \frac{2(2x)^5}{15} + \dots = 2x + \frac{8x^3}{3} + \frac{64x^5}{15} + \dots$ $1 - \cos 2x = 1 - (1 - \frac{(2x)^2}{2!} + \frac{(2x)^4}{4!} - \dots) = 2x^2 - \frac{2x^4}{3} + \dots$ Now,consider the numerator term $(	an 2x - 2 	an x)$: $	an 2x - 2 	an x = (2x + \frac{8x^3}{3} + \frac{64x^5}{15} + \dots) - 2(x + \frac{x^3}{3} + \frac{2x^5}{15} + \dots)$ $= (2x - 2x) + (\frac{8}{3} - \frac{2}{3})x^3 + (\frac{64}{15} - \frac{4}{15})x^5 + \dots = 2x^3 + 4x^5 + \dots$ So,$(	an 2x - 2 	an x)^2 = (2x^3 + 4x^5 + \dots)^2 = 4x^6 + \dots$ The numerator is $x^2(	an 2x - 2 	an x)^2 = x^2(4x^6 + \dots) = 4x^8 + \dots$ The denominator is $(1 - \cos 2x)^4 = (2x^2 - \frac{2x^4}{3} + \dots)^4 = (2x^2)^4 + \dots = 16x^8 + \dots$ Taking the limit: $\lim _{x ightarrow 0} \frac{4x^8}{16x^8} = \frac{4}{16} = \frac{1}{4}$

$\lim _{x \rightarrow 0} \frac{x^2(\tan 2 x-2 \tan x)^2}{(1-\cos 2 x)^4}=$

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