I have proven that $e^x > x^3$ for $x>5$, can I prove that $\lim \frac{x^3}{e^x} = 0$? In order to calculate the limit
$$\lim_{x\to\infty} \frac{x^3}{e^x} = 0$$
I've verified that:
$$f(x) = e^x-x^3\\f'(x) = e^x-3x^2\\f''(x) = e^x-6x\\f'''(x) = e^x-6$$
Note that $x>3 \implies f'''(x)>0$, therefore $f''(x)$ is crescent. Then, I find that $x = 3 \implies e^3-6\cdot3 >0$. Then, $x>3 \implies f'(x)$ is crescent. I just have to find an $x$ for $f'(x)$ such that $f'(x)>0$, then $f$ will be crescent. It happens that $x=4 \implies f'(x)>0$. Then, $x>4 \implies f(x)>0$. But for $x = 5$, $f(x)>0$, and $f$ is crescent, then:
$$x>5 \implies e^x-x^3>0\implies e^x>x^3$$
Therefore, some limits can be calculated easely, like:
$$x>5 \implies e^x>x^3\implies\lim_{x\to\infty}e^x>\lim_{x\to\infty}x^3 = \infty$$
But for the limit I want, we have:
$$x>5 \implies e^x>x^3\implies 1>\frac{x^3}{e^x}\implies \frac{x^3}{e^x}<1$$
I cannot, therefore, simply prove that this limit is equal to $0$, just by this inequality.
Any ideas?
 A: Hint: show that $e^x > x^4$ for $x$ sufficiently large. Then $\frac{x^3}{e^x}<\frac1x$. 
A: $$f(x)=e^x-x^3=>f'(x)=e^x-3x^2=>f''(x)=e^x-6x=>f'''(x)=e^x-6$$ Now $f'''(x)\gt 0$ for $x\gt3$. Therefore $f''(x)$ is concave after $x\gt3$. Similarly as you showed $f'(x)\gt 0$ for $x\gt4$. Hence you see that the slope of $f(x)$ ,i.e ,$f'(x)$ increases without any bounds after $x\gt 4$. So, at $x=\infty$, slope$=\infty$. This graphically means that when x is very large, $e^x\gt\gt\gt x^3$. Hence $$lim_{x\to \infty} \frac{x^3}{e^x}=0$$ 
A: $$\lim_{x \to 0} \frac{x^3}{e^x} = \lim_{x \to 0} \frac{x^3}{1 + x + x^2/2! + x^3/3! + \dots} = \frac 01 = 0$$
And
$$\begin{align}\lim_{x \to \infty} \frac{x^3}{e^x} 
  &= \lim_{x \to \infty} \frac{x^3}{1 + x + x^2/2! + x^3/3! + x^4/4!+\dots} 
\\&= \lim_{x \to \infty} \frac{1}{x^{-3} + x^{-2} + x^{-1}/2! + 1/3! + x/4!+\dots} \\&= \frac{1}{\infty}
\\&= 0
\end{align}$$
A: From L'Hospital's rule ($x\rightarrow\infty$):
$$\lim\frac{x^3}{e^x}=\lim\frac{3x^2}{e^x}=\lim\frac{6x}{e^x}=\lim\frac{6}{e^x}=0$$

Maybe from the squeeze theorem (if you can't use L'Hospital).
If $x\rightarrow \infty$ then $x^3>1$. So $\frac{x^3}{e^x}\ge \frac{1}{e^x}$
If $x\rightarrow \infty$ then $e^x \ge x^5$ (similar to your proof)
$$\frac{x^3}{x^5}\ge\frac{x^3}{e^x}\ge \frac{1}{e^x}$$
A: here is another way to show $\lim_{x \to infty}\frac{x^3}{e^x} = 0$ assuming the said limit exists, which can be proven once you establish $\frac{x^3}{e^x}$ is positive and decreasing for $x > 1.$ 
suppose $\lim_{x \to \infty} \frac{x^3}{e^x} = L.$ now replace $x$ by $2x$ limit should still be the same, that is,  $L = \lim_{x \to \infty}\frac{8x^3}{(e^x)^2} = 8L\lim_{x \to \infty}\frac{1}{e^x} = 0.$ 
