What is the limit of $((x-1)/x)^x$ as $x$ goes to infinity?

Question

I came across this limit when considering bootstrapping since $1 - ((x-1)/x)^x$ is the probability that observation $j$ from a sample of size $x$ does appears in a bootstrap sample also of size $x$. In other words, one minus this quantity is the probability of picking $j$ when choosing $x$ elements from $\{ 1, \ldots, x\}$ with replacement.

Using a computer, this limit appears to converge to approximately $0.3678611$ (this is the value for $x=10000$ to 7 d.p). I thought to try applying L'Hôpital's rule, knowing that the derivative of $x^x$ is $x^x(ln(x) + 1)$, but I could not make any progress.

So $(1+1/x)^x$ is the standard limit for $e$, I'm not seeing how to get to $(1-1/x)^x$ — Chris Russell, Apr 29 '21 at 13:30
@ChrisRussell In fact $\lim (1+\frac1u)^u=e$ here consider $u=-x$ so we have $\lim ((1-\frac1x)^{-x})^{-1}=(e)^{-1}$ — Etemon, Apr 29 '21 at 13:32
@Soheil, This substitution is inappropriate to use, since we are using $\lim_{x\to\infty}$ and $\lim_{u\to\infty}$. — Martund, Apr 29 '21 at 14:05
Does this answer your question? Proof for $e^z = \lim \limits_{x \rightarrow \infty} \left( 1 + \frac{z}{x} \right)^x$ — Martund, Apr 29 '21 at 14:10
I don't like citing results as though they are magic trick but ... $\lim_{x\to \infty} (1 +\frac k x)^x = e^k$. SO if $k = -1$ abracadabra $\lim_{x\to \infty} (\frac {x-1}x)^x = e^{-1} = \frac 1e$. ... uh... I guess it's not so bad if I take Soheil's comment. But $u=\frac xk$ then $\lim (1+\frac kx)^x = \lim(1 +\frac 1u)^{ku} = e^k$..... — fleablood, Apr 29 '21 at 16:15

score 1 · Accepted Answer · edited Apr 29 '21 at 13:58

1

So you have $$ \left(\frac{x-1}{x}\right)^x = \left(\frac{x}{x-1}\right)^{-x} = \left(1+\frac{1}{x-1}\right)^{-x} \to e^{-1} $$

edited Apr 29 '21 at 13:58

Chris Russell

244

answered Apr 29 '21 at 13:27

gt6989b

54,422

What is the limit of $((x-1)/x)^x$ as $x$ goes to infinity?

1 Answers1