No two sided limit exists

Question

I'm trying to calculate the limit of

$ \lim_{x \to 0} \frac{(e^x -1)}{(x^2)}$

I tried L'Hospital and apparently as it diverges and twice applied you get $\frac{1}{2}$. This is wrong apparently. How can I stop this mistake from happening, what can I spot beforehand?

Answer 1

Let $f(x)=e^x$

Then $\lim_{x \to 0} \frac{e^x-1}{x}=f'(0)=1$

Using this you can conclude that the side limits are different thus the limit does not exist

Answer 2

If you apply L'hospital once you get

$$\lim_{x\to 0} \frac{e^x}{2x}.$$

Since the numerator of this goes to $1$, you can't apply L'hospital a second time.