Indeterminate Quotients $\displaystyle\frac{0}{0}$ or
$\displaystyle\frac{\pm\infty}{\pm\infty}$
We start with indeterminate quotients, where either the numerator
and denominator are both going to zero or are both going to
infinity (positive or negative in any combination).
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 $0 \cdot \pm\infty$
Here we consider indeterminate products where one factor is going
to zero and the other is going to $\pm\infty$. Notice: we cannot apply L'Hospital's
rule to a product -- we must manipulate the product to get a
quotient before applying the rule.