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?  
 A: Simply observe that:
$$\frac{x^2}{e^x-1}=x\frac{x}{e^x-1}\to0\cdot 1=0$$
NOTE 
in this case algebric rule for multiplication holds
NOTE
$$\frac{x}{e^x-1}=\frac{1}{\frac{e^x-1}{x}}\to \frac{1}{1}=1$$
A: Let $f(x)=e^x$. 
Then $f(0)=1$ thus $$\lim_{x \to 0} \frac{e^x-1}{x}=\lim_{x \to 0}\frac{f(x)-f(0)}{x-0}=f'(0)=1$$
So we have: $$\lim_{x \to 0} \frac{x}{e^x-1}=\frac{1}{f'(0)}=1$$
Continue from here..
A: Just for fun, if you assume the limit exists, here are a couple of ways to show it must be zero.


*

*If $L=\lim_{x\to0}{x^2\over e^x-1}$, then, letting $u=2x$, which also tends to $0$ as $x$ tends to $0$, we have


$${x^2\over e^x-1}={e^x+1\over4}{4x^2\over(e^x+1)(e^x-1)}={e^x+1\over4}{(2x)^2\over e^{2x}-1}={e^x+1\over4}{u^2\over e^u-1}\to{1+1\over4}L={1\over2}L$$
so $L={1\over2}L$, which implies $L=0$.


*For $x\gt0$ we have $e^x\gt1$, so ${x^2\over e^x-1}\gt0$, hence $\lim_{x\to0}{x^2\over e^x-1}\ge0$. But for $x\lt0$ we have $e^x\lt1$, so ${x^2\over e^x-1}\lt0$, hence $\lim_{x\to0}{x^2\over e^x-1}\le0$.  


Note, neither of these prove that the limit is $0$, they just say it can't be any other real number.
A: $e^{x}=1+x+\frac{x^2}{2}+\frac{x^3}{3!}+O(x^4)$, so:
$$\frac{x^2}{e^x-1}=\frac{x^2}{1+x+\frac{x^2}{2}+\frac{x^3}{3!}+O(x^4)-1}=\frac{x^2}{x+\frac{x^2}{2}+\frac{x^3}{3!}+O(x^4)}=\frac{x}{1+x+\frac{x}{2}+\frac{x^2}{3!}+O(x^3)}\to\frac{0}{1}=0$$
A: Let $f:x\to e^x$
$\displaystyle\lim_{x\to 0}\; x\cdot\dfrac{x}{e^x-1}=\lim_{x\to 0}\; x\cdot\dfrac{x-0}{e^x-e^0}=\lim_{x\to 0}\; x\dfrac{1}{f'(0)}=\lim_{x\to 0}\; x\dfrac{1}{e^0}=0$
A: If you know that
$e^x \ge 1+x$,
then
$e^x-1 \ge x$
so
$\dfrac{x^2}{e^x-1}
\le \dfrac{x^2}{x}
= x
\to 0$
as
$x \to 0$.
A: $$
\begin{aligned}
\lim _{x \rightarrow 0} \frac{x^{2}}{e^{x}-1} &=\lim _{x \rightarrow 0} \frac{x}{\frac{e^{x}-1}{x}} \\
&=\frac{\displaystyle \lim _{x \rightarrow 0} x}{\displaystyle \lim _{x \rightarrow 0} \frac{e^{x}-1}{x}} \\
&=\frac{0}{1} \\
&=0
\end{aligned}
$$
