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I want to show for all $n\in\mathbb N$ $$\lim_{x\rightarrow\infty}\frac{x^{n}}{\exp(x^2)}=0$$

I am pretty sure that I have to use L'Hospital. I've tried induction:

$n=1$: $$\lim_{x\rightarrow\infty}\frac x{\exp(x^2)}=\lim_{x\rightarrow\infty}\frac1{2x\exp(x^2)}=0$$

And for $n\rightarrow n+1$: $$\lim_{x\rightarrow\infty}\frac{x^{n+1}}{\exp(x^2)}=\lim_{x\rightarrow\infty}\frac{(n+1)x^n}{2x\exp{x^2}}$$

And now I am stuck. The term $2x$ really annoys my for my induction hypothesis.

Any hints?

sheldoor
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  • Do you know the power series for the exponential? – Potato May 30 '13 at 22:29
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    $\frac 12$ can be pulled out of the limit. $x$ in the bottom cancels with one of $x$'s in the top, so you end up with $\frac{n+1}2\lim_{x\to\infty}x^{n-1}\exp(-x^2)$. Then by the induction hypothesis ($\lim_{x\to\infty}x^{k}\exp(-x^2) = 0$ for all $k \le n$), you get $0$. – Tunococ May 30 '13 at 22:29
  • @Tunococ you mean $k\leq n$, right? ;) – sheldoor May 30 '13 at 22:33
  • @sheldoor Thanks :) – Tunococ May 30 '13 at 22:34
  • You could also use $ \ \lim_{x \rightarrow \infty} \ x^n e^{-x} = 0 \ $ together with the "squeeze theorem" (since both functions are bounded below by zero) to obtain your limit. – colormegone May 30 '13 at 22:39
  • Why be annoyed at the $2x$, it is trying to help. By induction hypothesis, $\lim_{n\to\infty}\frac{(n+1)x^n}{e^{x^2}}=0$. So since for $x\gt 1/2$, we have $\frac{(n+1)x^n}{2xe^{x^2}}\lt \frac{(n+1)x^n}{e^{x^2}}$, by Squeezing the limit is $0$. – André Nicolas May 30 '13 at 23:17

5 Answers5

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From where you left it:

$$\lim_{x\to\infty}\frac{(n+1)\color{red}{x^n}}{2\color{red} xe^{x^2}}=\frac{n+1}2\lim_{x\to\infty}\frac{x^{n-1}}{e^{x^2}}\stackrel{\text{Inductive Hyp.}}=\frac{n+1}2\cdot 0=0$$

DonAntonio
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For $ \ x > 1 \ , $

$$ 0 \ \le \ e^{-x^2} \ \le \ e^{-x} $$

and also $ \ 0 \ < \ x^n \ \text{for} \ n \in \mathbb{N}$ , so

$$ 0 \ \le \ x^ne^{-x^2} \ \le \ x^ne^{-x} \ .$$

It is much easier to show $ \ \lim_{x \rightarrow \infty} \ x^ne^{-x} \ = \ 0 \ $ through l'Hopital's Rule. Thence, the "Squeeze Theorem" leads to

$$ 0 \ \le \ \lim_{x \rightarrow \infty} x^ne^{-x^2} \ \le \ \lim_{x \rightarrow \infty} x^ne^{-x} \ \Rightarrow \ \lim_{x \rightarrow \infty} x^ne^{-x^2} \ = \ 0 \ . $$

colormegone
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$$\lim_{x\rightarrow\infty}\frac{x^{n}}{\exp(x^2)}=\left( \lim_{x\rightarrow\infty}\frac{x}{\exp(x^2/n)}\right)^n=0$$

Now, you only need to apply L'H once.

Second solution

$$\lim_{x\rightarrow\infty}\frac{x^{n}}{\exp(x^2)}=\lim_{x\rightarrow\infty}\frac{\exp(n \ln x)}{\exp(x^2)}=\lim_{x\rightarrow\infty}\exp(n \ln x-x^2)=0$$

N. S.
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Note that for positive $x$, $x^{2n}\le e^{x^2}$ (by considering the taylor series), so $\frac{x^n}{e^{x^2}}\le\frac{x^n}{x^{2n}}=\frac{1}{x^n}$, hence $\lim\limits_{x\to\infty}\frac{x^n}{e^{x^2}}\le\lim\limits_{x\to\infty}\frac{1}{x^n}=0$.

JLA
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Applying L'Hospital $n$ times would do the trick too (assuming $n$ is fixed):

$$ \lim_{x\to\infty}\frac{x^{n}}{e^{x^2}} = \lim_{x\to\infty}\frac{(x^{n})'}{(e^{x^2})'} = \lim_{x\to\infty}\frac{nx^{n-1}}{e^{x^2} 2x}, $$ again we have an $\infty/\infty$ type limit, repeating L'Hospital until we get a constant in the denominator: $$ \lim_{x\to\infty}\frac{(x^{n})^{(n)}}{(e^{x^2})^{(n)}} = \lim_{x\to\infty}\frac{n!}{e^{x^2} p(x)} = 0, $$ where $p(x)$ is a degree $n$ polynomial in $x$.

Shuhao Cao
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