Limit $\lim_{x\rightarrow 0}\frac{e^{x}-1}{x^{2}}$ and problems The problem is straight forward enough, but I'm stuck.
$$\lim_{x\rightarrow 0} \frac{e^x-1}{x^2}$$
Since I get $0 / 0$ I can use L'Hopital's rule, but after the first differentiation I get $1/0$.  So, I can't continue with using that rule.  What approach next?
 A: Here's a fun, unconventional way to think about it.  If we suppose the limit exists, we have
\begin{align}
\lim\limits_{x\to 0} \frac{e^x-1}{x^2}&=\lim\limits_{x\to 0} \frac{e^x-1}{x}\cdot\frac1x\\
&=\lim\limits_{x\to 0} \frac{e^x-1}{x}\cdot\lim\limits_{x\to 0}\frac1x\\
&=1\cdot\lim\limits_{x\to 0}\frac1x
\end{align}
But we know that for $\lim\limits_{x\to 0}\frac1x$, the limit does not exist.  Curious.
A: You have correctly identified that you can apply L'Hopital's rule once to find 
$$\lim_{x\rightarrow 0} \frac{e^x}{2x}$$
Next consider the left-hand limit
$$\lim_{x\rightarrow 0^{-}} \frac{e^x}{2x}=\frac{1}{0^{-}}=-\infty$$
and the right-hand limit
$$\lim_{x\rightarrow 0^{+}} \frac{e^x}{2x}=\frac{1}{0^{+}}=\infty$$
so that
$$-\infty=\lim_{x\rightarrow 0^{-}} \frac{e^x}{2x}\neq \lim_{x\rightarrow 0^{+}} \frac{e^x}{2x}=\infty$$
What can you now conclude about the limit as $x\to 0$?
A: As noticed, recall that by standard limit


*

*$ \lim_{x\rightarrow 0}\frac{e^x-1}{x}= 1$
or by definition of derivative with $f(x)=e^x$


*

*$ \lim_{x\rightarrow 0}\frac{e^x-1}{x}=\lim_{x\rightarrow 0}\frac{f(x)-f(0)}{x-0}=f(0)=1$
therefore
$$\frac{e^x-1}{x^2}=\frac1x \cdot\frac{e^x-1}{x}\to\pm \infty$$
