lim _{x rightarrow 0} frac{x tan 4x - 2x tan | vedclass

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(B) We use the Taylor series expansion for $	an 	heta = 	heta + \frac{	heta^3}{3} + \frac{2	heta^5}{15} + \dots$ and $1 - \cos 	heta = 2 \sin^2(

Vedclass · Accepted Answer

$\frac{1}{4}$

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(B) We use the Taylor series expansion for $	an 	heta = 	heta + \frac{	heta^3}{3} + \frac{2	heta^5}{15} + \dots$ and $1 - \cos 	heta = 2 \sin^2(\frac{	heta}{2}) \approx 2(\frac{	heta}{2})^2 = \frac{	heta^2}{2}$.The denominator is $(1 - \cos 4x)^2 \approx (\frac{(4x)^2}{2})^2 = (8x^2)^2 = 64x^4$.The numerator is $x(4x + \frac{(4x)^3}{3} + \dots) - 2x(2x + \frac{(2x)^3}{3} + \dots)$.$= (4x^2 + \frac{64x^4}{3}) - (4x^2 + \frac{16x^4}{3}) = \frac{48x^4}{3} = 16x^4$.Thus,the limit is $\lim _{x ightarrow 0} \frac{16x^4}{64x^4} = \frac{16}{64} = \frac{1}{4}$.

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

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