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I am wondering how to prove $$\lim_{x\to \infty} \frac{e^x}{x^n}=\infty$$

I was thinking of using L'Hospital's rule? But then not sure how to do the summation for doing L'Hospital's rule n times on the denominator? Or whether it would be easier using longs like $\lim_{x\to \infty} \ln(e^x)-\ln(x^n)$?

Thank you!

Bernard
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  • You can use induction on $n$. Start by proving that $\lim_{x\to\infty}\frac{e^x}{1}=\infty$. Then, Assume that $\lim_{x\to\infty}\frac{e^x}{x^{n-1}}=\infty$. This implies that $\lim_{x\to\infty}\frac{e^x}{nx^{n-1}}=\infty$. By L'Hospital it follows that $\lim_{x\to\infty}\frac{e^x}{x^n}=\infty$. – user647486 Apr 05 '19 at 20:48
  • Taylor Series Expansion of $e^x$? – mrtaurho Apr 05 '19 at 20:48
  • Bernard.Thanks for editing.Deleted, Lord Shark was a bit faster. – Peter Szilas Apr 05 '19 at 21:07

6 Answers6

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One trick is to note that $$e^x=1+x+\frac{x^2}2+\cdots+\frac{x^{n+1}}{(n+1)!}+\cdots>\frac{x^{n+1}}{(n+1)!}$$ for $x>0$, so that $$\frac{e^x}{x^n}>\frac{x}{(n+1)!}$$ which of course tends to $\infty$ with $x$.

Angina Seng
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  • Thank you! Understood this, haven't learnt about power/taylor series yet but this makes sense :) –  Apr 05 '19 at 21:02
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You can certainly use L'Hopital's $n$ times. That is, for each $n\geq 0$ we have $$\lim_{x\to\infty}\frac{e^x}{x^n}=\lim_{x\to\infty}\frac{e^x}{nx^{n-1}}=\cdots=\lim_{x\to\infty}\frac{e^x}{n!}=\infty$$ since at each stage we are in $\frac{\infty}{\infty}$ indeterminate form.

Dave
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You can use L'Hospital. But, the key thing to notice the following. The derivatives of $x^n$ in ascending order are

$$nx^{n-1}, n(n-1)x^{n-2}, n(n-1)(n-2)x^{n-3},..., n!x, n!$$

Any $k$-th derivative for $k<n$ is going to have a limit of $\infty$ as $x\to\infty$. Also, any derivative of $e^x$ is $e^x$. Therefore, by applying L'Hospital successively, you will always have the form $\frac{\infty}{\infty}$, meaning that you can apply L'Hospital yet again. Eventually, you will have $n!$ in the denominator and the numerator will remain $e^x$, giving

$$\lim\limits_{x\to\infty} \frac{e^x}{n!}$$

Haris Gušić
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Using logs:

$\exp\left(\log e^x-\log x^n\right)=\exp\left(x-n\log x\right)=\exp\left(x\left(1-n\frac{\log x}{x}\right)\right)\to\exp\infty(1)=\exp(\infty)=\infty$

Lorenzo B.
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You just need to know that $$ \lim_{t\to\infty}\frac{e^t}{t}=\infty \tag{*} $$ Indeed $$ \frac{e^x}{x^n}=\left(\frac{e^{x/n}}{x}\right)^n=\frac{1}{n^n}\left(\frac{e^{x/n}}{x/n}\right)^n $$ How do you show (*)? Just one application of l'Hôpital suffices.

egreg
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Your idea of L'Hopital's rule will work also, but you have to apply it $n$ times. You should get $$\lim_{x\to \infty} \frac{e^x}{x^n} \stackrel{?}{=} \lim_{x\to \infty} \frac{e^x}{nx^{n-1}} \stackrel{?}{=} \lim_{x\to \infty} \frac{e^x}{n(n-1)x^{n-2}}\stackrel{?}{=} \cdots \stackrel{?}{=} \lim_{x\to \infty} \frac{e^x}{n!} = \infty.$$

However you have to justify each individual step (why L'Hopital's rule applies), and why the final limit $(e^x/n!)$ exists.

Caleb Stanford
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