# How to prove that $\lim_{n\rightarrow\infty}\frac{n^3x^2}{e^{nx}}$ converges to zero $\forall x\in[0;+\infty)$?

Show that $$\lim_{n\rightarrow\infty}\frac{n^3x^2}{e^{nx}}=0\ \ \ \forall x\in[0;+\infty)$$

The thing is I clearly understand that $$e^{nx}$$ grows faster than $$n^3x^2$$ but I cannot come up with any formal solution.

P.S. As for $$x=0$$, the answer is clear.

By ratio test for $$x>0$$ we obtain
$$\frac{(n+1)^3x^2}{e^{(n+1)x}}\frac{e^{nx}}{n^3x^2}=\left(\frac{n+1}{n}\right)^3\frac1{e^x}\to \frac1{e^x}<1$$
Use the fact that $$e^{nx} \geq \frac {n^{4}x^{4}} {4!}$$ (from the series expansion).
If $$x=0$$ it is trivial. If $$x>0$$ you can write
$$\frac{{n^3 x^2 }} {{e^{nx} }} = \frac{{n^3 x^3 }} {{e^{nx} }}\frac{1} {x} = \frac{{t^3 }} {{e^t }}\frac{1} {x}$$ where $$t=nx$$. As $$n \to +\infty$$ you have that $$t\to +\infty$$ and it is well known that $$\mathop {\lim }\limits_{t \to + \infty } \frac{{t^3 }} {{e^t }} = 0$$ thus $$\mathop {\lim }\limits_{t \to + \infty } \frac{{t^3 }} {{e^t }}\frac{1} {x} = \frac{1} {x}\mathop {\lim }\limits_{t \to + \infty } \frac{{t^3 }} {{e^t }} = 0$$