While we can't divide by zero, we can still meaningfully ask what happens to a
ratio $\displaystyle{\frac{f(x)}{g(x)}}$ when both $f(x)$ and $g(x)$ go to
zero. In fact, this sort of ratio appears every time we take a derivative:
$$ f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}.$$
We can't just plug in $h=0$, since then both the numerator and denominator
would be zero. It shouldn't be too surprising that derivatives can help us evaluate limits
that look like $0/0$.

L'Hospital's Rule:
If $f(x)$ and $g(x)$ are continuous functions
with $f(a)=g(a)=0$, then
$$\lim_{x \to a}\, \frac{f(x)}{g(x)} = \lim_{x \to a}\, \frac{f'(x)}{g'(x)}.$$

There are other variations of this rule for evaluating limits that
look like $\infty/\infty$, $0 \cdot \infty$, $\infty - \infty$,
$1^\infty$ or $0^0$, or for taking limits as $x \to \pm \infty$.
All these limits are called
indeterminate forms.
The complete statement of L'Hospital's rule and its proof is at the end of
this learning module. For now we'll concentrate on learning how to
apply the rule with some examples.

Indeterminate Quotients: $0/0$ or $\infty/\infty$

We start with indeterminate quotients, where either the numerator and
denominator are both going to zero or are both blowing up.

Solution: First, note that both $f(x)=x$ and $g(x)=1-e^x$ are zero when
$x=0$. So L'Hospital's rule applies. Then
$$ \lim_{x \to 0} \frac{x}{1-e^x} = \lim_{x\to 0} \frac{1}{-e^x}
= \frac{1}{-e^0} = \frac{1}{-1} = -1.$$
Note that we are computing the ratio of $f'(x)=1$ and $g'(x)=-e^x$, not
the derivative of $f/g$!

Indeterminate Products: $0 \cdot \infty$

Here we consider indeterminate products where one factor is going
to zero and the other is blowing up.