limits without lhopital - Find $lim_{x to 0} frac{e^{2x}-1}{tan x}$

I'd like help finding

$$\lim_{x \to 0} \frac{e^{2x}-1}{\tan x}$$

without the use of L'Hôpital's rule.

So far I did this:

$$\lim_{x \to 0} \frac{e^{2x}-1}{\tan x}$$
$$=\lim_{x \to 0} \frac{\cos x(e^{2x}-1)}{\sin x}$$
$$=\lim_{x \to 0} \frac{x\cos x(e^{2x}-1)}{x\sin x}$$
$$=\lim_{x \to 0} \frac{\cos x(e^{2x}-1)}{x}$$

Basically I got nowhere. Any hints or partial solutions to help me?

By the way, the is the 1969 AP BC Multiple Choice #28.