Here's another slightly different approach using L'Hopital.
\(\displaystyle \L\\\lim_{x\to\0}\frac{1-cos(x)}{x^{2}}\)
Take derivative of numerator and denominator.
\(\displaystyle \L\\(1-cos(x))dx=sin(x)\)
\(\displaystyle \L\\x^{2}dx=2x\)
\(\displaystyle \L\\\lim_{x\to\0}\frac{1}{2}\frac{sin(x)}{x}\)
\(\displaystyle \L\\\frac{1}{2}\lim_{x\to\0}\frac{sin(x)}{x}\)
We could stop here because of the familiar limit(as Soroban showed), but I will trudge onward.
Now, do L'Hopital again by taking derivatives of numerator and denominator:
\(\displaystyle \L\\sin(x)dx=cos(x)\) and
xdx=1
\(\displaystyle \L\\\frac{1}{2}\lim_{x\to\0}(cos(x))\)
\(\displaystyle \H\\\frac{1}{2}\)