# How to prove that $\lim_{x\rightarrow \infty}\dfrac{x^2}{e^x}=0$?

I need to prove that $\lim_{x\rightarrow \infty}\dfrac{x^2}{e^x}=0$.

• Are you allowed to use l'Hôpital's theorem? – egreg Nov 13 '14 at 14:14
• Do you know about l'Hopital's rule? – Ritz Nov 13 '14 at 14:14
• Do you know that $\lim_{x\to\infty}\frac{x}{e^x}=0$? – Hagen von Eitzen Nov 13 '14 at 14:15
• @Ritz Not yet... – qexi Nov 13 '14 at 14:24
• @Hagen von Eitzen Yes – qexi Nov 13 '14 at 14:24

We have the well-known inequality $e^t\ge 1+t$ for all $t\in\mathbb R$. Therefore $$e^x=(e^{x/3})^3\ge \left(1+\frac x3\right)^3=1+x+\frac13x^2+\frac1{27}x^3$$ for $x\ge 0$, whence $$0\le\frac{x^2}{e^x}\le \frac{x^2}{1+x+\frac13x^2+\frac1{27}x^3}\to 0.$$
$$0 \leq \frac{x^2}{e^x}= \frac{x^2}{(e^{x/4})^4}\leq \left(\frac{x}{(1+x/4)^2}\right)^2.$$
For $x>0$, we have $e^x \geq \dfrac{x^3}{3!}$ (Taylor expansion of $e^x$). Hence, we have $$\dfrac{x^2}{e^x} \leq \dfrac{6}x \to 0 \text{ as }x \to \infty$$