Evaluate a limit requiring two applications of L'Hopital's Rule

Substituting $x=0$ gives $\frac{e^0 - 1 - 0}{0^2} = \frac{0}{0}$, an indeterminate form, so L'Hopital's Rule applies. Differentiating numerator and denominator separately gives $\frac{d}{dx}(e^x - 1 - x) = e^x - 1$ and $\frac{d}{dx}(x^2) = 2x$, so the limit becomes $\lim_{x\to 0} \frac{e^x - 1}{2x}$.

Checking this new limit at $x=0$ gives $\frac{e^0-1}{0} = \frac{0}{0}$, still indeterminate, so L'Hopital's Rule can be applied again. Differentiating again gives $\frac{d}{dx}(e^x-1) = e^x$ and $\frac{d}{dx}(2x) = 2$, so the limit becomes $\lim_{x\to 0} \frac{e^x}{2}$. Substituting $x=0$ now gives $\frac{e^0}{2} = \frac{1}{2}$, which is the value of the original limit.