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$$\lim_{x\to\infty} \bigg(\frac{x-1}{x}\bigg)^x = \frac{1}{e} $$

Is there any comfortable way to see that this is true?

The following thing occured to me: take log and use taylor series for $\log(1+x)$: $$\log(1-1/x)^x = x\log(1-1/x) = -1-\frac{1}{2x}-\frac{1}{3x^2}-... $$

If we take the $\lim_{ x\to\infty}$ it is pretty clear all the terms go to zero except the first one. This if log(something) is -1 then this something = $e^{-1} = 1/e$

Seems to work but doesn't help me "feel" it.

Appreciate your help/feedback!

snoram
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1 Answers1

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One pretty common definition of $e$ is $$e=\lim_{n \to \infty}\left(1+\frac{1}{n}\right)^n$$ Further, if you have a sequence $(x_n)$ such that $x_n \to x \neq 0$ then you also know $$\frac{1}{x_n}\to \frac{1}{x}$$ Now obviously $e \neq 0$ so $$\frac{1}{e} = \lim_{n \to \infty} \frac{1}{\left(1+\frac{1}{n}\right)^n} \\ = \lim_{n \to \infty}\left(1+\frac{1}{n}\right)^{-n} \\ = \lim_{n \to \infty}\left(\frac{n+1}{n}\right)^{-n} \\ = \lim_{n \to \infty}\left[\left(\frac{n+1}{n}\right)^{-1}\right]^n \\ = \lim_{n \to \infty}\left(\frac{n}{n+1}\right)^n$$ Your definition of $e$ is slightly different, but you should be able to make a very similar argument to get the same result.

graydad
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