Why $\lim_{n \to \infty}(1-1/n)^n = e^{-1}$?

Question

I am curious why

$$\lim_{n \to \infty} \left(1-\frac{1}{n}\right)^n = e^{-1}.$$

I can see why $$\lim_{n \to \infty} \left(1+\frac{m}{n}\right)^n = e^m$$ for some $m > 0$ by substituting $k = n/m$. But why does the result (Im guessing) also hold for the case $m <0$?

Answer 1

$$f(x)=\left(1-\frac{1}{x} \right)^x$$ Change of variable : $$x=\frac{1}{\epsilon}\quad\to\quad f(x)=g(\epsilon)=\exp\left(\frac{\ln(1-\epsilon)}{\epsilon} \right)$$ $\frac{\ln(1-\epsilon)}{\epsilon}=\frac{1}{\epsilon}\left(-\epsilon -\frac{\epsilon^2}{2}-\frac{\epsilon^3}{3}-...\right)=-1-\frac{\epsilon}{2}-\frac{\epsilon^2}{3}-...$

$x\to\infty \qquad \epsilon \to 0 \qquad \frac{\ln(1-\epsilon)}{\epsilon} \to -1$ $$f(x\to\infty)=g(\epsilon\to 0)\to \exp(-1)$$

Answer 2

Verify that for $n \ge 2$ we have

$(1- \frac{1}{n})^n=\frac{1}{(1+ \frac{1}{n-1})^{n-1}} \frac{1}{1+ \frac{1}{n-1}}$.

Answer 3

$$\left(1-\frac{1}{n}\right)^{-n}=\left(\frac{n}{n-1}\right)^{n}=\left(\frac{k+1}{k}\right)^{k}\left(\frac{k+1}{k}\right)=\left(1+\frac{1}{k}\right)^k\left(1+\frac{1}{k}\right)\to e$$

where we made the change of variable $n=k+1$

Answer 4

Another observation $$\lim_{n \to \infty}(1+1/n)^n \times\lim_{n \to \infty}(1-1/n)^n = \\ \lim_{n \to \infty}(1+1/n)^n (1-1/n)^n =\\ \lim_{n \to \infty} (1-\frac{1}{n^2})^n \to 1$$ now we know $\lim_{n \to \infty} \left(1+\frac{1}{n}\right)^n = e^{+1}.$ so

$$\lim_{n \to \infty}(1+1/n)^n \times\lim_{n \to \infty}(1-1/n)^n =1\\ e^{+1}\times \lim_{n \to \infty}(1-1/n)^n =1\\ \to \lim_{n \to \infty}(1-1/n)^n =\frac{1}{e}$$