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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.

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By ratio test for $x>0$ we obtain

$$\frac{(n+1)^3x^2}{e^{(n+1)x}}\frac{e^{nx}}{n^3x^2}=\left(\frac{n+1}{n}\right)^3\frac1{e^x}\to \frac1{e^x}<1$$

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Use the fact that $e^{nx} \geq \frac {n^{4}x^{4}} {4!}$ (from the series expansion).

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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 $$ \mathop {\lim }\limits_{t \to + \infty } \frac{{t^3 }} {{e^t }}\frac{1} {x} = \frac{1} {x}\mathop {\lim }\limits_{t \to + \infty } \frac{{t^3 }} {{e^t }} = 0 $$

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