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

Question

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

Answer 1

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 $$0\le\frac{x^2}{e^x}\le \frac{x^2}{1+x+\frac13x^2+\frac1{27}x^3}\to 0.$$

Answer 2

For $x>0$, we have $e^x \geq \dfrac{x^3}{3!}$ (Taylor expansion of $e^x$). Hence, we have $$\dfrac{x^2}{e^x} \leq \dfrac{6}x \to 0 \text{ as }x \to \infty$$

Answer 3

$$0 \leq \frac{x^2}{e^x}= \frac{x^2}{(e^{x/4})^4}\leq \left(\frac{x}{(1+x/4)^2}\right)^2.$$