Proof: $\lim_{n\to\infty} (1-\frac{1}{n})^{-n} = e$

Question

If $e:=\lim_{n\to\infty} (1 + \frac{1}{n})^n$, prove that $$ \lim_{n\to\infty} \left(1-\frac{1}{n}\right)^{-n} = e $$ without using the property that says: if $\lim_{n\to\infty} a_n = \infty$, $\lim_{n\to\infty} x_n = x$, then $\lim{n\to\infty}(1+\frac{x_n}{a_n})^{a_n} = e^x$...

I've tried rewriting $1-\frac{1}{n}$, and operating $\lim_{n\to\infty}(1-\frac{1}{n})^n = 1 / \lim_{n\to\infty}(1+\frac{1}{n})^n $ but I couldn't prove it. Any hint? Thanks in advance.

Answer 1

\begin{gather} \lim\limits_{n\to\infty}\left(1-\dfrac{1}{n}\right)^{-n}=\lim\limits_{n\to\infty}\left(\dfrac{n-1}{n}\right)^{-n}= \\ =\lim\limits_{n\to\infty}\left(\dfrac{n}{n-1}\right)^{n}=\lim\limits_{n\to\infty}\left[\left(1+\dfrac{1}{n-1}\right)^{n-1}\left(1+\dfrac{1}{n-1}\right)\right]=e \end{gather}

Answer 2

Note that $$\left(1-\frac{1}{n}\right)\left(1+\frac{1}{n}\right)=1-\frac{1}{n^2}$$

And $$1\geq \left(1-\frac 1{n^2}\right)^n \geq 1-\frac{n}{n^2}$$

The last is an application of Bernoulli's inequality, with $x=-\frac{1}{n^2}$ and $r=n$.

So $(1-\frac 1{n^2})^n\to 1$ and hence $(1-1/n)^n\to e^{-1}$. and $(1-1/n)^{-n}\to e$

Answer 3

We have $$\lim_{n\to\infty}\left(1-\frac{1}{n}\right)^{-n}=\lim_{n\to\infty}\left(1+\frac{(-1)}{n}\right)^{n\cdot(-1)}.$$ Now, we can bring the limit inside the $(\cdot)^{-1}$ to have $$\left[\frac{1}{e}\right]^{-1}=e.$$ Observe that I used what you have written so that $$\lim_{n\to\infty}\left(1-\frac{1}{n}\right)^n=\frac{1}{e}.$$ For the proof of the last equality, see @amWhy's comment.