# Prove that $\lim_{x\to\infty} \frac{e^x}{x^n}=\infty$ without using L'hôpital's rule

I know how to prove that $\lim_{x\to\infty} \frac{e^x}{x^n}=\infty$ using LHR. I'm trying to show it without using it, but it's not going very well.

Using the power series of $e^x$ is also not allowed.

• Can we use the power series for $e^x$? Easy then. – André Nicolas Feb 20 '14 at 23:00
• Take the power series of $e^x$. See that $e^x > \dfrac{x^{n+1}}{(n+1)!}$ for $x > 0$. – Daniel Fischer Feb 20 '14 at 23:01
• I forgot to mention that it's not allowed to use the power series... I know how to apply it though. Iv'e edited my post. – Galc127 Feb 20 '14 at 23:03

An inductive argument on how fast they grow shouldn't be too hard to do.

Here are two algebraic variations:

$$\lim_{x \to +\infty} \frac{e^x}{x^n} = \left( \lim_{x \to +\infty} \frac{e^{x/n}}{x} \right)^n = \exp \left(\lim_{x \to +\infty} x - n \ln x \right)$$

• Is there a method of seeing $x-n\ln x\to\infty$ which is inherently simpler than $\frac{e^x}{x^n}\to\infty$? – Jonathan Y. Feb 20 '14 at 23:17
• @Jonathan Y.: It follows from knowing that $x$ grows asymptotically faster than $\ln x$; if that's something one knows right off, then that would make for a quick proof. A rewriting of that limit that might also be useful is $$\ln x \left( \frac{x}{\ln x} - n \right)$$ – user14972 Feb 20 '14 at 23:20
• I do believe that's quite the same thing as knowing $e^x$ grows asymptotically faster than $x^n$, is it not? – Jonathan Y. Feb 20 '14 at 23:21
• @Jonathan: They are all very similar. Different people work better with and remember different things than other people, though, or have different ideas triggered by seeing the different forms. If all I knew is how $e^x$, $x$, and $\ln x$ compared in growth, I think I'd find the last of the three the easiest to work with. – user14972 Feb 20 '14 at 23:22

HINT: Show that $\lim\limits_{x\to\infty}\dfrac{\ln x}x = 0$ by comparing the integral $\displaystyle\int_1^x \dfrac{dt}t$ to, say, $\displaystyle\int_1^x \dfrac{dt}{\sqrt t}$.

$e=1+\alpha$ with $\alpha > 0$. Let $k>n$, $m=\left[x\right]$ (largest integer not greater than $x$), it follows that

$$e^x>e^m=(1+\alpha)^m=\sum_{i=0}^m {m \choose i}\,\alpha^i>{m \choose k}\,\alpha^k$$ for $m \geq k$. For $m>2k$ we have $${m\choose k}\,\alpha^k=\frac{m(m-1)\ldots(m-k+1)}{k!}\,\alpha^k>\left(\frac{m}{2}\right)^k\frac{\alpha^k}{k!}$$ and $x \leq m+1$, $x^n\leq(m+1)^n$, hence $$\frac{e^x}{x^n}> \frac{\alpha^k}{2^k k!}\left(\frac{m}{m+1}\right)^k(m+1)^{n-k}>\frac{\alpha^k}{4^kk!}m^{n-k}$$