Finding the limit of $(1-\cos x)/x^2$ $$\lim _{x \to 0}{1-\cos x\over x^2}={2\sin^2\left(\frac{x}{2}\right)\over x^2}={\frac{2}{x^2}\cdot {\sin^2\left(\frac{x}{2}\right)\over \left(\frac{x}{2}\right)^2}}\cdot\left(\frac{x}{2}\right)^2$$
now $$\lim_{x \to 0}{\sin^2\left(\frac{x}{2}\right)\over \left(\frac{x}{2}\right)^2}=\left({\sin\left(\frac{x}{2}\right)\over \left(\frac{x}{2}\right)}\right)^2=1^2$$
So we have $$\frac{2}{x^2}\cdot \left(\frac{x}{2}\right)^2=\frac{2}{x^2}\cdot \left(\frac{x^2}{4}\right)=\frac{1}{2}$$
Are the moves right?
 A: I'm surprised nobody used L'Hospital's Rule:
$$\lim _{x \to 0}{1-\cos x\over x^2}\equiv \lim _{x \to 0}{\sin x\over 2x}\equiv\lim _{x \to 0}{\cos x\over 2}=\frac{1}{2}
$$
A: 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:
$$
\lim_{x\to0}\frac{1-\cos x}{x^2}=
\lim_{x\to0}\frac{1-\cos^2 x}{x^2(1+\cos x)}=
\lim_{x\to0}\frac{1}{1+\cos x}\left(\frac{\sin x}{x}\right)^{\!2}
$$
A: To provide a correction to your own work I would remove the $\lim$ at first because I want to simplifies to the maximum the expression and at the last the computation, as follows:
$${1-\cos x\over x^2}={2\sin^2\left(\frac{x}{2}\right)\over x^2}={\frac{2}{x^2}\cdot {\sin^2\left(\frac{x}{2}\right)\over \left(\frac{x}{2}\right)^2}}\cdot\left(\frac{x}{2}\right)^2 ={\sin^2\left(\frac{x}{2}\right)\over \left(\frac{x}{2}\right)^2}\cdot \frac{1}{2}$$ therefore
$$\lim{1-\cos x\over x^2}=\lim{\sin^2\left(\frac{x}{2}\right)\over \left(\frac{x}{2}\right)^2}\cdot \frac{1}{2}=1\cdot\frac{1}{2}=\frac{1}{2}.$$ 
