Answer
$\dfrac{1}{2}$
Work Step by Step
Consider: $\lim\limits_{x \to 0}f(x)=\lim\limits_{x \to 0}\dfrac{1-\cos x}{x^2}$
Need to check that the limit has an indeterminate form.
Thus, $f(0)=\dfrac{1-\cos 0}{0}=\dfrac{0}{0}$
Now, apply L-Hospital's rule such as: $\lim\limits_{a \to b}f(x)=\lim\limits_{a \to b}\dfrac{g'(x)}{h'(x)}$
This implies:
$\lim\limits_{x \to 0}\dfrac{\sin x}{2x}=\dfrac{1}{2} \lim\limits_{x \to 0}\dfrac{\sin x}{x}=\dfrac{1}{2} \lim\limits_{x \to 0}\dfrac{\cos x}{1}=\dfrac{1}{2}$
