Are there any situations in which L'Hopital's Rule WILL NOT work? Today was my first day learning L'Hopital's Rule, and I was wondering if there are any situations in which you cannot use this rule, with the exception of when a limit is determinable. 

Here's a similar example to Ittay Weiss's, but more nefarious, since in Ittay's example the limit was easily seen to be equal to 1:
$$\lim_{x \rightarrow 0}\frac{e^{-\frac{1}{x^2}}}{x}$$
Both the numerator and denominator go to zero. But applying l'Hospital gives:
$$\lim_{x \rightarrow 0}\frac{2e^{-\frac{1}{x^2}}}{x^3}$$
And applying l'Hospital again:
$$\lim_{x \rightarrow 0}\frac{4e^{-\frac{1}{x^2}}}{3x^5}$$
We see that the situation is getting worse, not better!
The way out is to forget about l'Hospital entirely, and try to replace $e^{-1/x^2}$ with something more manageable. We have, for all $x$:
$$e^x \geq x$$
And so for all $x > 0$:
$$e^{-x} \leq \frac{1}{x}$$
Hence, for all $x \neq 0$:
$$e^{-\frac{1}{x^2}} \leq x^2$$
This implies:
$$0 \leq \frac{e^{-\frac{1}{x^2}}}{x} \leq \frac{x^2}{x} = x \qquad (x > 0)$$
$$x = \frac{x^2}{x} \leq \frac{e^{-\frac{1}{x^2}}}{x} \leq 0 \qquad (x < 0)$$
And the squeeze theorem tells us that the original limit is zero.
