$\lim _{x\to 0}\left(\frac{e^x-1}{\:x^3}\right)$ I apply L'Hopital thrice and I get 1/6 but here they've stopped at infinity.

What is the correct answer?
Note that L’Hopital rule is used only for limits of $\frac00$ or $\frac{\infty}{\infty}$ form. 

Right now, on substituting $x=0$, we get a $\frac00$ form. But, if we differentiate it once, we get: $$\lim_{x \to 0} \frac{e^x}{3x^2}$$ which is not of $\frac00$ form. So, we cannot use L’Hopital rule here.

Evaluating the above limit, the answer is obviously $\infty$.
I don't know how you used L’Hopital rule but look(assume I'm taking the limit all the time):

$$\frac{e^x-1}{x^3}=\frac{e^x}{3x^2}$$

This is a limit in the form of $\frac1{0}$, which in this case is determinate form, it goes to $+\infty$ because $\lim_{x\to0}x^2$  goes to $0$ from the positive direction

The answer is infinity