Here's a "naturally occurring" example:
The Taylor expansion of
$f(x) = \exp\left( -\frac{1}{x^2}\right)$ $f(0) = 0$
is
$0 + 0x + 0x^2 + 0x^3 + 0x^4 + ...$
The simplest way I know to obtain these coefficients is to prove by induction that the $n$th derivative of $f(x)$, for $x \neq 0$, is a rational function times $\exp\left( -\frac{1}{x^2}\right)$ (the rational function will vary with $n$), a form that you'd likely guess if you tried computing the first two or three derivatives. It is not very hard to prove the induction step, namely that the $(n+1)$st derivative is a rational function times $\exp\left( -\frac{1}{x^2}\right)$ when $x \neq 0$. Using this result, we can now prove (again, by induction) that each $n$th derivative evaluated at $x = 0$ exists and is equal to $0$. This time, in proving the inductive step, which involves evaluating the $(n+1)$st derivative at $x = 0$, you're going to need L'Hopital's rule to evaluate the limit as $h \rightarrow 0$ of
$\frac{f^{(n)}(0 + h) - f^{(n)}(0)}{h},$
where you will be assuming that $f^{(n)}(0 + h) = f^{(n)}(h) = R(h) \cdot \exp\left( -\frac{1}{h^2}\right)$ for some rational function $R(h)$. However, if you start "L'Hopital differentiating" right away (with respect to $h$), you'll find things will spiral out of control. (Try it by using $R(h) = \frac{P(h)}{Q(h)}$.) To fix this, make the variable change $u = \frac{1}{h}$ and then take the limit as $u \rightarrow \infty$ of the $u$-version, using L'Hopital's rule.