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$\lim _{x \rightarrow 0} \frac{8}{x^8}\left[\begin{array}{l}1-\cos \left(\frac{x^2}{2}\right)-\cos \left(\frac{x^2}{4}\right) \\ +\cos \left(\frac{x^2}{2}\right) \cdot \cos \left(\frac{x^2}{4}\right)\end{array}\right]$ is equal to
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$\frac{1}{32}$
$\lim _{x \rightarrow 0} \frac{8}{x^8}\left(1-\cos \left(\frac{x^2}{2}\right)-\cos \left(\frac{x^2}{4}\right)+\cos \left(\frac{x^2}{2}\right) \cdot \cos \left(\frac{x^2}{4}\right)\right)$
$\lim _{x \rightarrow 0} \frac{8}{x^8}\left\{1\left(1-\cos \frac{x^2}{2}\right)-\cos \left(\frac{x^2}{4}\right)^2 \cdot\left(1-\cos \left(\frac{x^2}{2}\right)\right)\right\}$
$\lim _{x \rightarrow 0} \frac{8}{x^8}\left\{\left(1-\cos \frac{x^2}{2}\right)\left(1-\cos \left(\frac{x^2}{4}\right)\right)\right\}$
$\left[\because(1-\cos \theta)=2 \sin ^2 \frac{\theta}{2}\right.$
$\lim _{x \rightarrow 0} \frac{8}{x^8}\left\{\left(2 \sin ^2\left(x^2 / 4\right)\right)\left(2 \sin ^2\left(x^2 / 8\right)\right\}\right.$
$32$.$\lim _{x \rightarrow 0} \frac{1}{x^8}\left(\frac{\sin \left(\frac{x^2}{4}\right)}{\frac{x^2}{4}}\right)^2 \times \frac{x^4}{16} \times\left(\frac{\sin \left(x^2 / 8\right)}{x^2 / 8}\right)^2 \times \frac{x^4}{64}$
$\lim _{x \rightarrow 0} \frac{32 x^8}{16 \times 64 x^8}\left(\frac{\sin x^2 / 4}{x^2 / 4}\right)^2\left(\frac{\sin x^2 / 8}{x^2 / 8}\right)^2$
$=\frac{1}{32} \cdot 1 \cdot 1$ $\left\{\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right\}$
$=\frac{1}{32}$
$\lim _{x \rightarrow 0} \frac{8}{x^8}\left\{1\left(1-\cos \frac{x^2}{2}\right)-\cos \left(\frac{x^2}{4}\right)^2 \cdot\left(1-\cos \left(\frac{x^2}{2}\right)\right)\right\}$
$\lim _{x \rightarrow 0} \frac{8}{x^8}\left\{\left(1-\cos \frac{x^2}{2}\right)\left(1-\cos \left(\frac{x^2}{4}\right)\right)\right\}$
$\left[\because(1-\cos \theta)=2 \sin ^2 \frac{\theta}{2}\right.$
$\lim _{x \rightarrow 0} \frac{8}{x^8}\left\{\left(2 \sin ^2\left(x^2 / 4\right)\right)\left(2 \sin ^2\left(x^2 / 8\right)\right\}\right.$
$32$.$\lim _{x \rightarrow 0} \frac{1}{x^8}\left(\frac{\sin \left(\frac{x^2}{4}\right)}{\frac{x^2}{4}}\right)^2 \times \frac{x^4}{16} \times\left(\frac{\sin \left(x^2 / 8\right)}{x^2 / 8}\right)^2 \times \frac{x^4}{64}$
$\lim _{x \rightarrow 0} \frac{32 x^8}{16 \times 64 x^8}\left(\frac{\sin x^2 / 4}{x^2 / 4}\right)^2\left(\frac{\sin x^2 / 8}{x^2 / 8}\right)^2$
$=\frac{1}{32} \cdot 1 \cdot 1$ $\left\{\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right\}$
$=\frac{1}{32}$
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