$\lim _{x\to 0}\left(\frac{e^x-1}{\:x^3}\right)$ 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 $\infty$.