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$\lim _{x \rightarrow 0} \frac{x \tan 4 x-2 x \tan 2 x}{(1-\cos 4 x)^2}=$
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The correct answer is:
$\frac{1}{4}$
$\lim _{x \rightarrow 0} \frac{x \tan 4 x-2 x \tan 2 x}{(1-\cos 4 x)^2}$
$=\lim _{x \rightarrow 0} \frac{x\left[(4 x)+\frac{1}{3}(4 x)^3+\frac{2}{15}(4 x)^5+\ldots\right] -2 x\left[(2 x)+\frac{1}{3}(2 x)^3+\frac{2}{15}(2 x)^5+\ldots\right]}{\left(2 \sin ^2 2 x\right)^2}$
$=\lim _{x \rightarrow 0} \frac{x\left[\frac{1}{3} 2^4 x^3\left(2^2-1\right)+\frac{2}{15} 2^6 x^5\left(2^4-1\right)+\ldots\right]}{2^2\left(\frac{\sin 2 x}{2 x}\right)^4(2 x)^4}$
$=\frac{\frac{1}{3} 2^4\left(2^2-1\right)}{2^2 \times 2^4}=\frac{1}{4}$
$=\lim _{x \rightarrow 0} \frac{x\left[(4 x)+\frac{1}{3}(4 x)^3+\frac{2}{15}(4 x)^5+\ldots\right] -2 x\left[(2 x)+\frac{1}{3}(2 x)^3+\frac{2}{15}(2 x)^5+\ldots\right]}{\left(2 \sin ^2 2 x\right)^2}$
$=\lim _{x \rightarrow 0} \frac{x\left[\frac{1}{3} 2^4 x^3\left(2^2-1\right)+\frac{2}{15} 2^6 x^5\left(2^4-1\right)+\ldots\right]}{2^2\left(\frac{\sin 2 x}{2 x}\right)^4(2 x)^4}$
$=\frac{\frac{1}{3} 2^4\left(2^2-1\right)}{2^2 \times 2^4}=\frac{1}{4}$
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