Compute $\lim_{x\to0}{1 - e^{-x}\over e^x - 1}$ 
Compute $\displaystyle \lim_{x\to0}{1 - e^{-x}\over e^x - 1}$.

So first and foremost, I know what the answer to this question is:
$$\begin{align}\lim_{x\to0}{1 - e^{-x}\over e^x - 1} &= \lim_{x\to0}\left({1 - e^{-x}\over e^x - 1}\right)\cdot{e^x\over e^x} \tag{$\star$}\\ & = \lim_{x\to0}{e^x - 1\over (e^x - 1)e^x} \\ &= \lim_{x\to0}{1 \over e^x} \\ &= {1\over e^0} \\ &= 1.\end{align}$$
Unfortunately, I found this solution by complete accident. I can't really justify the motivation of why I performed the step $(\star)$ other than "I saw the $e^{-x}$ and thought to try and get rid of it.
Initially the first thought that I had was to use the conjugate of one of the functions, but we end up not going anywhere with the problem (There's still that nasty $e^{-x}$ using either conjugate). Is this likely the intended solution to this problem? I unfortunately cannot see any other way of solving the problem.
 A: Here's three ways of doing it:
Taylor
You can use Taylor, i.e. $e^{x} \sim 1 + x$, so
$$\frac{1 - e^{-x}}{e^{x} - 1} \sim \frac{1 - (1-x)}{1+x-1} = \frac{x}{x} =1$$
Factorizing
Another (but equivalent) way is factor $e^{x}$ from the denominator
$$\displaystyle \lim_{x\to0}{1 - e^{-x}\over e^x - 1} = \displaystyle \lim_{x\to0}{1 - e^{-x}\over e^{x}(1 - e^{-x})}= \lim_{x\to0}\frac{1}{e^{x}} = \frac{1}{e^0} = 1 $$
L'Hopital
You can solve by L'Hopital, i.e.
$$\displaystyle \lim_{x\to0}{1 - e^{-x}\over e^x - 1} = \displaystyle \lim_{x\to0} \frac{e^{-x}}{e^{x}} = \frac{e^{0}}{e^{0}} = 1$$
You could use $(\epsilon,\delta)$ definition as well
A: You may also use the substitution $x=\ln y$. This may make your trick seem more natural:
$${1 - e^{-x}\over e^x - 1} = {1 - e^{-\ln y}\over e^{\ln y} - 1} ={1 - {1 \over y}\over y - 1} ={y - 1\over y(y - 1)} = { 1\over y}\stackrel{y \to 1}{\longrightarrow}1$$
A: Another way is writing $e^{-x}$ as $\dfrac{1}{e^x}$.
Then, $$\begin{align}\lim_{x\to0}{1 - e^{-x}\over e^x - 1} &= \lim_{x\to0}\left({1 - \dfrac{1}{e^{x}}\over e^x - 1}\right) \\ & = \lim_{x\to0}{e^x - 1\over (e^x - 1)e^x} \\ &= \lim_{x\to0}{1 \over e^x} \\ &= {1\over e^0} \\ &= 1\end{align}$$
See? Almost same as your solution. I think one can easily see through these steps.
A: My own approach would be to notice the similarity between the numerator and denominator, and by educated guess perform the factorization
$$e^x-1=e^x(1-e^{-x})$$
and the rest is easy.
Or to perform the substitution $t:=e^x$, giving
$$\frac{1-\dfrac1t}{t-1}=\frac1t.$$
