One minus cosine over x squared

Idea. Turn the cosine into a sine, so that we can use $\frac{\sin u}u\to1$.

Computation. From $\cos x=1-2\sin^2\frac x2$ we get
$$\frac{1-\cos x}{x^2}=\frac{2\sin^2\frac x2}{x^2}=\frac12\left(\frac{\sin\frac x2}{\frac x2}\right)^2.$$
As $x\to0$, $\frac x2\to0$ and $\frac{\sin(x/2)}{x/2}\to1$. So the limit is $\dfrac12\cdot1^2=\dfrac12$.

Remark. This tells us that $\cos x\approx1-\dfrac{x^2}2$ near $0$, the beginning of the Taylor expansion of cosine.