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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.

Please help.

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    $\begingroup$ Can we use the power series for $e^x$? Easy then. $\endgroup$ – André Nicolas Feb 20 '14 at 23:00
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    $\begingroup$ Take the power series of $e^x$. See that $e^x > \dfrac{x^{n+1}}{(n+1)!}$ for $x > 0$. $\endgroup$ – Daniel Fischer Feb 20 '14 at 23:01
  • $\begingroup$ 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. $\endgroup$ – Galc127 Feb 20 '14 at 23:03
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An inductive argument on how fast they grow shouldn't be too hard to do.

Here are two algebraic variations:

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

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    $\begingroup$ Is there a method of seeing $x-n\ln x\to\infty$ which is inherently simpler than $\frac{e^x}{x^n}\to\infty$? $\endgroup$ – Jonathan Y. Feb 20 '14 at 23:17
  • $\begingroup$ @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 $$ \ln x \left( \frac{x}{\ln x} - n \right) $$ $\endgroup$ – user14972 Feb 20 '14 at 23:20
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    $\begingroup$ I do believe that's quite the same thing as knowing $e^x$ grows asymptotically faster than $x^n$, is it not? $\endgroup$ – Jonathan Y. Feb 20 '14 at 23:21
  • $\begingroup$ @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. $\endgroup$ – user14972 Feb 20 '14 at 23:22
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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, $\displaystyle\int_1^x \dfrac{dt}{\sqrt t}$.

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$e=1+\alpha$ 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 $$ \frac{e^x}{x^n}> \frac{\alpha^k}{2^k k!}\left(\frac{m}{m+1}\right)^k(m+1)^{n-k}>\frac{\alpha^k}{4^kk!}m^{n-k} $$

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