Limit of $e^x/x^3$ at infinity without l'Hopital I'm looking for a nice proof that $$\lim_{x\to\infty} \frac{e^x}{x^3} \to +\infty$$
without derivatives (i.e., Taylor or l'Hopital).
My current approach is to change it into a sequence: For any $x$, there exists $n$ such that $n\leq x\leq n+1$. Then
  $$\frac{e^x}{x^3}\geq\frac{e^n}{(n+1)^3}=:a_n.$$
Since
  $$\lim_{n\to\infty} \frac{a_n}{a_{n-1}}
  = \lim_{n\to\infty} \frac{e^n}{(n+1)^3}\frac{n^3}{e^{n-1}}=e>1,$$
we get that
  $$\lim_{x\to\infty} \frac{e^x}{x^3} \geq \lim_{n\to\infty} a_n \to +\infty.$$
However, this approach seems quite complicated to me. Is there any elementary proof that were significantly simpler? For instance, do I overlook some trivial bound (i.e., not derived from Taylor) on $e^x$ that could be helpful?
PS: We define $e^x:=\lim\limits_{n\to\infty} (1+x/n)^n$.
PPS: I tried to check whether a question like this exists, but I couldn't find it. If it exists, I apologize.
 A: Use $x=3t$ so the limit is
$$
\lim_{t\to\infty}\frac{e^{3t}}{27t^3}=
\lim_{t\to\infty}\frac{1}{27}\left(\frac{e^t}{t}\right)^{\!3}
$$
Thus you see that you just need to show
$$
\lim_{x\to\infty}\frac{e^x}{x}=\infty
$$
Now the problem is in how you define $e^x$.
With $e^x=\lim_{n\to\infty}(1+x/n)^n$, the Bernoulli inequality gives
$$
\left(1+\frac{x}{n}\right)^{\!n}\ge 1+x
$$
but this seems to weak. It is not if you consider
$$
\frac{e^x}{x}=\frac{(e^{x/2})^2}{x}\ge
\frac{(1+x/2)^2}{x}
$$
Note that this technique shows that
$$
\lim_{x\to\infty}\frac{e^x}{x^\alpha}=\infty
$$
for every $\alpha>0$, just in one swoop: consider $e^x=(e^{x/\alpha})^\alpha$.
A: Let $x\gt 0$. Expand $\left(1+\frac{x}{n}\right)^n$ using the binomial theorem. We find that the term in $x^4$ is
$$\frac{n(n-1)(n-2)(n-3)}{4!n^4}x^4.$$
Note that if $n\ge 6$ then $n(n-1)(n-2)(n-3)\gt \frac{n^4}{2^3}$.
It follows that $e^x\gt \frac{1}{8\cdot 4!}x^4$, and therefore 
$$\frac{e^x}{x^3}\gt \frac{x}{8\cdot 4!}.$$  
A: There is an "elementary" inequality $e^t \geq t+1$, from which we can deduce $$\frac{e^x}{x^3} = \frac{(e^{x/4})^4}{x^3}  \geq \frac{(\frac{x}{4} + 1)^4}{x^3} \geq \frac{1}{4^4}\frac{x^4}{x^3} = \frac{x}{4^4} \to \infty$$.
A: Assuming you can use the definition of $e^x$.
$e^x$ always has all positive terms in the expansion for positive x.
There is always a term in the definition of $e^x$ of $\frac{x^4}{24}$ and so $e^x > \frac{x^4}{24}$
$$e^x=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+... > \frac{x^4}{24}$$
Above some value of $x$ we must have $x^4 > 24x^3$ and that's $x>24$
So at any value above $x=24$ we are guaranteed
$$\frac{e^x}{x^3} > x$$
And the limit of that as $x$ approaches $\infty$ is $\infty$
