how to evaluate $\lim_{x\to0} (x^2)/(e^x-1) $ without L'Hospital So I'm trying to evaluate limit written in the title without L'Hospital nor series (cause its not introduced in our course),I tried to use this recognized limit $\lim_{x\to0} \frac{\left(e^x−1\right)}x =1$
so our limit equal $\lim_{x\to0} x\left(\frac{x}{e^x−1}\right)$.
I'm not sure how to prove that $\left(\frac{x}{e^x−1}\right) = 1$ 
Any facts I can use here or other algebraic manipulation I can use to evaluate limit?  

$$
\begin{aligned}
\lim _{x \rightarrow 0} \frac{x^{2}}{e^{x}-1} &=\lim _{x \rightarrow 0} \frac{x}{\frac{e^{x}-1}{x}} \\
&=\frac{\displaystyle \lim _{x \rightarrow 0} x}{\displaystyle \lim _{x \rightarrow 0} \frac{e^{x}-1}{x}} \\
&=\frac{0}{1} \\
&=0
\end{aligned}
$$
