I am given the following limit to solve, presumably without knowledge of L'Hôpital's rule:
$$\lim_{x\to0}\left({\frac{x}{1-\cos x}}\right)^2$$
I tried using trigonometric identities (namely Pythagorean) to solve it, but with no luck.
Answer
$\displaystyle \lim_{x\to0}\left({\frac{x}{1-\cos x}}\right)^2=$
$\displaystyle = \lim_{x\to0}\left({\frac{x}{2\sin^2 \frac x2}}\right)^2=$
$\displaystyle = \lim_{x\to0}\left({\frac{ \frac{x}{2}}{\sin^2 \frac x2}}\right)^2=$
$\displaystyle = \lim_{x\to0}\left({\frac{ \frac{x}{2}}{\sin \frac x2}} \cdot \frac{1}{\sin \frac x2}\right)^2=$
$\displaystyle =\left(1\cdot \frac{1}{0} \right)^2=$
$=\infty$
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