Limit of $e^x/x^3$ at infinity without l'Hopital

Question

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.

Answer 1

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 $e^x=(e^{x/\alpha})^\alpha$.

Answer 2

There is an "elementary" inequality $e^t \geq t+1$, from which we can deduce $$\frac{e^x}{x^3} = \frac{(e^{x/4})^4}{x^3} \geq \frac{(\frac{x}{4} + 1)^4}{x^3} \geq \frac{1}{4^4}\frac{x^4}{x^3} = \frac{x}{4^4} \to \infty$$.

Answer 3

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 $$\frac{e^x}{x^3}\gt \frac{x}{8\cdot 4!}.$$

Answer 4

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 $\infty$