calculus - $lim_{xrightarrow 0^+}frac{(1+cos x)}{(e^x-1)}= infty$ using l'Hopital

I need to show $$\lim_{x\rightarrow 0^+}\frac{1+\cos x}{e^x-1}=\infty$$

I know that, say, if you let
$f(x) = 1 + \cos x$
and
$g(x) = \dfrac{1}{e^x-1}$,
and then multiply the limits of $f(x)$ and $g(x)$, you get $\frac{2}{0}$. I can't figure out how to make it work for l'Hopital's rule however, i.e. how to rewrite it so that it is in the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$.

I also tried multiplying $h(x)$ by the conjugate of $f(x)$, but I don't think this is fruitful. Any hints appreciated.