Finding the limit of $(1-\cos x)/x^2$

Correct, but too complicated (and missing several $\lim_{x\to0}$). $$ \lim_{x\to0}\frac{1-\cos x}{x^2}= \lim_{x\to0}\frac{2\sin^2(x/2)}{4(x/2)^2}= \lim_{x\to0}\frac{1}{2}\left(\frac{\sin(x/2)}{(x/2)}\right)^{\!2}= \frac{1}{2} $$

Alternative way: $$ \lim_{x\to0}\frac{1-\cos x}{x^2}= \lim_{x\to0}\frac{1-\cos^2 x}{x^2(1+\cos x)}= \lim_{x\to0}\frac{1}{1+\cos x}\left(\frac{\sin x}{x}\right)^{\!2} $$

I'm surprised nobody used L'Hospital's Rule:

$$\lim _{x \to 0}{1-\cos x\over x^2}\equiv \lim _{x \to 0}{\sin x\over 2x}\equiv\lim _{x \to 0}{\cos x\over 2}=\frac{1}{2} $$