Limit $\lim_{x\rightarrow 0}\frac{e^{x}-1}{x^{2}}$ and problems

Question

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?

Answer 1

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.

Answer 2

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 $x\to 0$?

Answer 3

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

$$\frac{e^x-1}{x^2}=\frac1x \cdot\frac{e^x-1}{x}\to\pm \infty$$