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

Substituting $x=0$ gives numerator $e^0 - 1 = 0$ and denominator $0$, so the limit has the indeterminate form $0/0$. This means L'Hopital's Rule applies, since both functions are differentiable near $x=0$.

Differentiate numerator and denominator separately: the derivative of $e^x - 1$ is $e^x$, and the derivative of $x$ is $1$. So the limit becomes $\lim_{x\to 0} \frac{e^x}{1} = e^0 = 1$.