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What is the limit $x\to+\infty$ of $x^n \cdot e^{-x}$ ? My guess is that it is $0$ because the exponential will go faster to $0$ than the $n$th power go to $+\infty$, but I don't see which theorem to apply to reach the result.

Thanks for any help !

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    $\begingroup$ Try l'Hopital's rule $n$ times. $\endgroup$ – Theo Bendit Jul 22 '15 at 8:46
  • $\begingroup$ Ugh, that sounds ugly. Deriving $\frac{1}{e^x}$ already duplicates the exponential. Isn't there a fully writable way ? $\endgroup$ – Lærne Jul 22 '15 at 8:51
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    $\begingroup$ Well, you have to write it first as $\frac{x^n}{e^x}$. Then differentiating the denominator is just differentiating $e^x$, which is extremely easy. (Also, for the record, differentiating $\frac{1}{e^x} = e^{-x}$ is not much more difficult.) $\endgroup$ – Theo Bendit Jul 22 '15 at 8:54
  • $\begingroup$ Oops... That the kind of stuff too easy to notice ^^. Thank you. $\endgroup$ – Lærne Jul 22 '15 at 8:56
  • $\begingroup$ Only one application of l'Hopital's rule is necessary if you take logarithms first. $\endgroup$ – Siméon Jul 22 '15 at 17:08
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This can be shown without using L'Hospital as follows:

Let $$\exp:\mathbb R\rightarrow\mathbb R,~x\mapsto\sum\limits_{n=0}^\infty \frac{1}{n!}x^n$$ and $P(x)$ a polynomial. We show that $$\lim\limits_{x\to\infty}\frac{P(x)}{\exp(x)}=0.$$

Let $n$ be the degree of $P$, then there exists $c>0,R>0$ with $|P(x)|\leq c|x|^n$ for alle $|x|\geq R$. From the series representation we directly obtain $$\exp(x)>\frac{1}{(n+1)!}x^{n+1}$$ for all $x\geq 0$. Using this we get $$\left|\frac{P(x)}{\exp(x)}\right|\leq \frac{cx^n}{\frac{1}{(n+1)!}x^{n+1}}=\frac{c(n+1)!}{x}$$ for all $x\geq R$. As $$\lim\limits_{x\to\infty}\frac{c(n+1)!}{x} =0$$ we are done.

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  • $\begingroup$ Yeah, I prefer to have a way without l'Hospital. Thank you ! $\endgroup$ – Lærne Jul 22 '15 at 9:01
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If you do not want to use L'Hopital's rule, this is another posibility. If $x>0$ then $$ e^x=\sum_{k=0}^\infty\frac{x^k}{k!}>\frac{x^{n+1}}{(n+1)!}\implies e^{-x}<\frac{(n+1)!}{x^{n+1}}. $$

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Theo Bendit gave the hint. Write $$A=\frac{x^n}{e^x}=\frac uv$$ Apply L'Hospital a first time $$\frac {u'}{v'}=\frac{nx^{n-1}}{e^x}$$ Repeat $$\frac {u''}{v''}=\frac{n(n-1)x^{n-2}}{e^x}$$ $$\frac {u'''}{v'''}=\frac{n(n-1)(n-2)x^{n-3}}{e^x}$$After $n$ steps, you have a constant in the numerator (which you even do not need to know that it is $n!$) and still the beautiful $e^x$ in the denominator.

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  • $\begingroup$ Yeah thank you for the clarification. Apparently I'm not able to notice that $e^{-x} = 1/e^x$ or to derive $1/e^{-x}$. $\endgroup$ – Lærne Jul 22 '15 at 9:01
  • $\begingroup$ You are welcome ! $e^{-x} = 1/e^x$ was the trick !! $\endgroup$ – Claude Leibovici Jul 22 '15 at 9:02
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"My guess is that ... because the exponential will go faster to $0$ than the nth power go to $\infty$". This guess is actually based on the fact that $\lim_{x \to \infty}x^{n}e^{-x} = 0$ for all $n > 0$ and not the other way round.

As easy way out is to put $e^{x} = t$ so that $x = \log t$ and then $x^{n}/e^{x} = (\log t)^{n}/t$. Further if $t > 1$ then $$\log t = 2n\log t^{1/2n} \leq 2n(t^{1/2n} - 1) < 2nt^{1/2n}$$ so that $$0 < \frac{\log t}{t^{1/n}}< \frac{2n}{t^{1/2n}}$$ Using squeeze theorem we get that $(\log t)/t^{1/n} \to 0$ as $t \to \infty$ and hence $$(\log t)^{n}/t = ((\log t)/t^{1/n})^{n} \to 0$$

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  • $\begingroup$ Why does log(t)=2n*log(t)^1/2n ? $\endgroup$ – Student number x Feb 19 '18 at 1:09
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    $\begingroup$ @Studentnumberx: do you know that $\log a^b=b\log a$? If yes then see what happens when we put $a=t^{1/2n},b=2n$. $\endgroup$ – Paramanand Singh Feb 19 '18 at 3:33
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One way is to use l'Hospital (see comment by @theo-bendit ), without any "calculation": $\exp{x}=\sum_{k=0}^\infty x^k/k! > \sum_{k=0}^{n+1} x^k/k!$ and then $x^n/\exp{x} < x^n / \sum_{k=0}^{n+1} x^k/k!$ . We divide both parts of the fraction by $x^n$ and see that the limit is basically the same as that of $1/x$ for large $x$.

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$x^n\cdot e^{-x} = e^{-x+n\ln x}$. We show: $-x + n\ln x \to -\infty$ when $x \to +\infty$. First choose $x > 3n$, then: $-x+n\ln x = -1+\displaystyle \int_{1}^x \left(\dfrac{n}{t} - 1\right)dt=-1+\displaystyle \int_{1}^{3n} \left(\dfrac{n}{t}-1\right)dt+\displaystyle \int_{3n}^x \left(\dfrac{n}{t} - 1\right)dt < C(n) -\dfrac{2}{3}\left(x-3n\right) \to -\infty$ as $x \to +\infty$. Thus: $-x+n\ln x \to -\infty \Rightarrow x^\cdot e^{-x} \to 0$. Here $C(n)$ is a constant.

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  • $\begingroup$ I got you up until the very last step when you put in the bound < $C(n) ...$, but how exactly did you manage to bound the integral by $\frac{2}{3}(3n - x)$? $\endgroup$ – Requiens Nov 30 '18 at 18:24
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The answer is $n!$. Bring $e^{-x}$ in denominator and differentiate $n$ times

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  • $\begingroup$ The answer is $0$, since with many applications of L'Hospital rule gives $\lim_{x\to\infty} n! \cdot e^{-x}$, as other have pointed out, which is $n! \cdot 0 = 0$, not $n!$ $\endgroup$ – Lærne Jul 23 '15 at 7:16
  • $\begingroup$ how come exp(-x)=infinity as x tends to infinity? $\endgroup$ – Chirag Palan Jul 23 '15 at 11:17
  • $\begingroup$ I'm trying to get the limit of $x^n$ times $\exp(-x)$. Using L'Hospital, I get the limit of $n!$ times $\exp(-x)$, read Claude Leibovici's answer & comments for why. And $\lim_{x\to+\infty} \exp(\mathbf{-}x) = 0$, so I have $n! \cdot 0 = 0$... Where do you see $\lim_{x\to+\infty}\exp(-x) = +\infty$ ? $\endgroup$ – Lærne Jul 23 '15 at 12:25
  • $\begingroup$ but you can not apply L hospital until it is (0/0) or (infinity/infinity) form, so to make it in that form you should send exp(-x) in denominator as exp(x) then it becomes the form where you apply L'hospihal rule. $\endgroup$ – Chirag Palan Jul 24 '15 at 9:25
  • $\begingroup$ Yes in that form we use L'hospital rule until we reach $\frac{n!}{e^x}$, which equals $n! \cdot e^{-x}$ which tends to $0$, not $n!$. $\endgroup$ – Lærne Jul 24 '15 at 14:20

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