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The value of $\lim _{x \rightarrow 0} \frac{1-\cos (1-\cos x)}{x^4}$ is
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Verified Answer
The correct answer is:
$\frac{1}{8}$
$\frac{1}{8}$
$1-\cos (1-\cos x)=2 \sin ^2\left(\frac{1-\cos x}{2}\right)$
$=2 \sin ^2\left(\sin ^2 \frac{x}{2}\right)$
$\therefore \lim _{x \rightarrow 0} \frac{1-\cos (1-\cos x)}{x^4}=\lim _{x \rightarrow 0} \frac{2 \sin ^2\left(\sin ^2 \frac{x}{2}\right)}{x^4}$
$=\lim _{x \rightarrow 0} \frac{2 \sin ^2\left(\sin ^2 \frac{x}{2}\right)}{\left(\sin ^2 \frac{x}{2}\right)^2} \times \frac{\sin ^4 \frac{x}{2}}{\left(\frac{x}{2}\right)^4 \times 16}$
$=\frac{1}{8}$
$=2 \sin ^2\left(\sin ^2 \frac{x}{2}\right)$
$\therefore \lim _{x \rightarrow 0} \frac{1-\cos (1-\cos x)}{x^4}=\lim _{x \rightarrow 0} \frac{2 \sin ^2\left(\sin ^2 \frac{x}{2}\right)}{x^4}$
$=\lim _{x \rightarrow 0} \frac{2 \sin ^2\left(\sin ^2 \frac{x}{2}\right)}{\left(\sin ^2 \frac{x}{2}\right)^2} \times \frac{\sin ^4 \frac{x}{2}}{\left(\frac{x}{2}\right)^4 \times 16}$
$=\frac{1}{8}$
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