Evaluating a 0/0 limit using L'Hopital's Rule

Substituting $x=0$ gives $\frac{1-\cos(0)}{0^2} = \frac{0}{0}$, which is indeterminate, so L'Hopital's Rule applies. Differentiating numerator and denominator separately: the derivative of $1-\cos(x)$ is $\sin(x)$, and the derivative of $x^2$ is $2x$. This gives the new limit $\lim_{x\to 0} \frac{\sin(x)}{2x}.$

Substituting $x=0$ into $\frac{\sin(x)}{2x}$ again gives $\frac{0}{0}$, so L'Hopital's Rule applies a second time. Differentiating again: the derivative of $\sin(x)$ is $\cos(x)$, and the derivative of $2x$ is $2$, giving $\lim_{x\to 0} \frac{\cos(x)}{2} = \frac{\cos(0)}{2} = \frac{1}{2}.$ This is no longer indeterminate, so the limit equals $\frac{1}{2}$.