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$\lim_{x\to\infty} \frac{e^x}{x^n}$

n is any natural number.

Using L'hopital doesn't make much sense to me. I did find this in the book:

"In a struggle between a power and an exp, the exp wins."

Can I refer that line as an answer? If the fraction would have been flipped, then the limit would be zero. But in this case I the limit is actually $\infty$

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    $\begingroup$ That statement you encountered is a nonrigorous version of a statement on growth rates; briefly, no matter how high you take $n$ in $x^n$, there is a value $x$ such that beyond it, $\exp(x)>x^n$. Now use this to see if a limit exists. $\endgroup$ – J. M. is a poor mathematician Nov 20 '10 at 14:06
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Repeated use of L'Hôpital's rule ($n$ times):

$\underset{x\rightarrow \infty }{\lim }\dfrac{e^{x}}{x^{n}}=\underset{% x\rightarrow \infty }{\lim }\dfrac{e^{x}}{nx^{n-1}}=\underset{x\rightarrow \infty }{\lim }\dfrac{e^{x}}{n(n-1)x^{n-2}}=\cdots =$ $=\cdots=\underset{x\rightarrow \infty }{\lim }\dfrac{e^{x}}{n(n-1)\cdots 3\cdot x^{2}}=\underset{x\rightarrow \infty }{\lim }\dfrac{e^{x}}{% n(n-1)\cdots 3\cdot 2x}=$ $=\underset{x\rightarrow \infty }{\lim }\dfrac{e^{x}}{% n(n-1)\cdots 3\cdot 2\cdot 1}=\underset{x\rightarrow \infty }{\lim }\dfrac{% e^{x}}{n!}=\infty$


Added:

To convince yourself: If you had $\underset{x\rightarrow \infty }{\lim }\dfrac{e^{x}}{x^{10}}$ you would have to apply L'Hôpital's rule ten times.

Added 2: Plot of $\dfrac{e^{x}}{x^{3}}$

alt text

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Use the fact that $e^x \ge \left( 1 + \frac{x}{n} \right)^n$ for any $n > 0$.

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HINT: One way of looking at this would be: $$\frac{1}{x^{n}} \biggl[ \biggl(1 + \frac{x}{1!} + \frac{x^{2}}{2!} + \cdots + \frac{x^{n}}{n!}\biggr) + \frac{x^{n+1}}{(n+1)!} + \cdots \biggr]$$

I hope you understand why i put the brackets inside those terms.

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  • $\begingroup$ Found a nice theorem that says something about this! $\endgroup$ – Algific Nov 20 '10 at 14:18
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Why isn't L'Hôpital a good solution? Just use induction.

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    $\begingroup$ L'Hôpital is a thing whose use should be avoided, as usually the alternatives are much more instructive. $\endgroup$ – J. M. is a poor mathematician Nov 20 '10 at 14:16
  • $\begingroup$ Considering the boy is just doing homework, I don't think a highbrow mathematical approach is needed here. Besides, I don't know what is so bad about L'Hôpital. Particularly in this case. $\endgroup$ – Raskolnikov Nov 20 '10 at 14:18
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I believe the problem is tailor-made for repeated application of L'Hopital's Rule, but here are some thoughts ...

You could note that $e^{x} = (e^{x/n})^n$, and consider $\left( \lim \frac{e^{x/n}}{x}\right)^n$, so that you are comparing an exponential to a single power of $x$, which might be a bit less daunting for you.

A bit more cleanly, and to make the numerator and denominator match better, define $y := \frac{x}{n}$. Then $$\frac{e^{x}}{x^n}=\frac{e^{ny}}{(ny)^n}=\frac{\left(e^{y}\right)^n}{n^n y^n}=\frac{1}{n^n}\frac{\left(e^{y}\right)^n}{y^{n}}=\frac{1}{n^n}\left(\frac{e^y}{y}\right)^n$$

Since $n$ is a constant, you can direct your limiting attention to $\frac{e^y}{y}$ (as $y \to \infty$, of course).

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Let's consider the limit to infinity when taking log of that expression:

$$\lim_{x\rightarrow\infty} \ln{\frac{e^x}{x^n}} = \lim_{x\rightarrow\infty}x-n\ln(x)=\lim_{x\rightarrow\infty}x(1-\frac{n\ln(x)}{x})=\infty$$

Therefore, the limit is $\infty$.

The proof is complete.

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