Is the limit $$\lim_{n→0} {n!\over{n^n}} $$ equal to $1$?
My proof attempt.
$${0!\over{0^0}} =\frac 1 1 =1.$$
Am I right? Please correct me if I am wrong.
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7What is your definition of $n!$ for non-integer values of $n$? Are you computing a one-sided limit $n\to 0+$? – Gary Apr 27 '22 at 07:17
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Does this answer your question? What's the limit of the sequence $\lim\limits_{n \to\infty} \frac{n!}{n^n}$? – JohannesPauling Apr 27 '22 at 07:21
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Suppose n!/n^n is a sequence a_n – roar de voir Apr 27 '22 at 07:21
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2@roardevoir Are you sure it is not $n\to \infty$? – Gary Apr 27 '22 at 07:22
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3@JohannesPauling Apparently here $n\to 0$ and not $n\to \infty$. – Gary Apr 27 '22 at 07:23
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The denominator tends to $1$ (the expression is only defined for positive real numbers) and the usual extenstion of the factorial $n!$ is $\Gamma(n+1)$ , in which case the numerator tends to $\Gamma(1)=1$ – Peter Apr 27 '22 at 07:24
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You should make explicit reference to limit laws. Other than that it's all good. – Thomas Anton Apr 27 '22 at 07:26
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1$n\rightarrow \infty$ would not give limit $1$ , but $0$ – Peter Apr 27 '22 at 07:29
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@Gary Oh, I misread the question. – JohannesPauling Apr 27 '22 at 07:31
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3For a seqence $(a_n){n=0}^{\infty}$ it may or may not make sense to talk about $\lim{n \to 0} a_n$, depending on your precise definition of limit. But with the definition that (probably) is the most commonly accepted, it's meaningless to even talk about $\lim_{x \to x_0} f(x)$ if $x_0$ is not an accumulation point of the the domain $D_f$, and then the limit in your question doesn't exist. – Hans Lundmark Apr 27 '22 at 08:04
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Not sure why my "$0+0$" was re-edited. It is crucial that $n$ ($x$ would be better anyway) approaches from the right. For negative values near $0$, the expression is not defined. For $n=0$ itself , it is also undefined , so we cannot just insert $n=0$ (in which case the exercise would be trivial). – Peter Apr 27 '22 at 08:04
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@HansLundmark Very good comment ! – Peter Apr 27 '22 at 08:05
1 Answers
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Saying that $0^0 = 1$ is wrong in general, and also $x^x$ is not clearly define for $x<0$, I assume you mean the limit $$ \lim_{x\to 0^+} \dfrac{x!}{x^x} $$ Moreover you have to define what $n!$ is when $n$ is not an integer.
Using the gamma function as extension of the factorial for non-integer: $$ \Gamma(z):= \int_0^{+\infty} t^{z-1}\cdot e^{-t}\;dt$$ For each $n\in\mathbb{N}$: $\Gamma(n+1) = n!$, in particular (useful later) $\Gamma(1) = 0! = 1$.
Then knowing that $\Gamma(z)$ is continuous for $z>0$ and that $\lim_{x\to 0^+} x^x = 1$ (see: Stack Question ), you get
$$ \lim_{x\to 0^+} \dfrac{x!}{x^x} = \lim_{x\to 0^+} \dfrac{\Gamma(x+1)}{x^x} = \dfrac{\lim_{x\to 0^+} \Gamma(x+1)}{\lim_{x\to 0^+} x^x} = \dfrac{\Gamma(1)}{1} = 1. $$
Leo
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In the end it turns out as a sophisticated way to validate the OP's proof attempt! – Gribouillis Apr 27 '22 at 08:13
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1"Saying that $0^0=1$ is wrong in general" I strongly disagree. Naively swapping out $x^y$ with $0^0$ in a limit $x, y\to 0$ is wrong in general. But I know of absolutely no issue with stating that $0^0 = 1$ always holds. Which is why I am of the conviction that $0^0 = 1$ always holds. You might disagree with me that $0^0 = 1$ always holds, and I'm more or less fine with that, although I, like everyone else, wish that my personal conventions were more universal. But I think you will find it difficult to say I am wrong. – Arthur Apr 27 '22 at 09:06
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@Arthur Despite my harsh words, I kind of agree with you. $0^0 = 1$ seems natural in many fields. Still, in the context of calculus and especially when working with limits (precisely the context of the question), we can't make a direct jump from $0^0$ to 1. In a different context (e.g. combinatorics, algebra, etc.) I would usually define $0^0$ as 1 (if need be) before working with it. Without context I like to leave $0^0$ as undefined. – Leo Apr 27 '22 at 09:43
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In my calculus book, sure we can make a direct jump from $0^0$ to $1$ when working with limits. What we can't do is make a direct jump from $\lim_{x, y\to 0}x^y$ to $0^0$. Discontinuities exist, and we are capable of working with them everywhere else, I don't see how it would be a problem in this particular case. Admittedly, "$0^0$" is also used to denote this common "indeterminate form", and this notational overload is unfortunate. On the other hand, Taylor expansions get messy if you don't allow for $0^0 = 1$, so even inside the field of calculus, most people use it at least some times. – Arthur Apr 27 '22 at 10:25