convergence divergence - Limit of the function $lim_{xto 0}(frac{e^{-x}-1}{x})$

I'm trying to solve this limit without the use of L'Hospital, but I'm doing something wrong. The limit should be:

$$\lim_{x\to 0}\left(\frac{e^{-x}-1}{x}\right) = -1$$

My attempted proof:
$$\lim_{x\to 0}(\frac{e^{-x}-1}{x}) = \lim_{n\to \infty}(\frac{e^{-(\frac{1}{n})}-1}{\frac{1}{n}}) = \lim_{n\to \infty}(\frac{n \cdot ( e^{-(\frac{1}{n})}-1)}{1}) = \lim_{n\to \infty}(\frac{n \cdot ( e^{0}-1)}{1}) = \lim_{n\to \infty}(\frac{0}{1}) = 0$$

I assume the mistake is that I've used the continuity of the $exp$ function.

Answer

A variation of Bernard's answer:

$$\frac{e^{-x}-1}x=\frac{1-e^x}{xe^x}=-\frac1{e^x}\cdot\frac{e^x-1}x\xrightarrow[x\to0]{}-\frac11\cdot(e^x)'_{x=0}=-e^0=-1$$