limit of $\left( 1-\frac{1}{n}\right)^{n}$

Question

limit of $$\left( 1-\frac{1}{n}\right)^{n}$$

is said to be $\frac{1}{e}$ but how do we actually prove it?

I'm trying to use squeeze theorem

$$\frac{1}{e}=\lim\limits_{n\to \infty}\left(1-\frac{1}{n+1}\right)^{n}>\lim\limits_{n\to \infty}\left( 1-\frac{1}{n} \right)^{n} > ??$$

Answer 1

Alternately $$e^x=\lim \left(1+\frac{x}{n}\right)^n$$ put $x=-1$

Answer 2

It easily follows from $(1+\frac1n)^n \longrightarrow e$.

Infact, $(1-\frac1n)^n=(\frac{n-1}{n})^n=(\frac{n}{n-1})^{-n}=(1+\frac1{n-1})^{-n} \longrightarrow e^{-1}$

Answer 3

Using your approach,

$$\left(1-\frac{1}{n+1}\right)^{n}>\left( 1-\frac{1}{n} \right)^{n}>\left( 1-\frac{1}{n} \right)^{n+1}$$

and

$$\lim\limits_{n\to \infty}\left(1-\frac{1}{n+1}\right)^{n}={1\over e}$$

$$ \lim\limits_{n\to \infty}\left( 1-\frac{1}{n} \right)^{n+1}=\lim\limits_{n\to \infty}\left( 1-\frac{1}{n} \right)^{n-1}\cdot\lim\limits_{n\to \infty}\left( 1-\frac{1}{n} \right)^{2}=\lim\limits_{n\to \infty}\left( 1-\frac{1}{n} \right)^{n-1} \cdot 1$$

$$= \lim\limits_{n-1\to \infty} \frac{1}{ \left( 1+\frac{1}{n-1} \right)^{n-1} } = \frac{1}{ \lim\limits_{n-1\to \infty} \left( 1+\frac{1}{n-1} \right)^{n-1} } = {1\over e}$$

Hence by squeeze theorem you know

$$\lim\limits_{n\to \infty}\left(1-\frac{1}{n}\right)^{n}={1\over e}$$

As discussed in the comment, your way of writing is not formally correct and you cannot use the squeeze theorem in one line.

Answer 4

I believe you just want the proof. I am not using squeeze theorem in this answer because you haven't said that you want to prove it only using squeeze theorem. This is a limit of the form $1^{\infty}$ which can be easily done by using the following steps( Here $\lim_{x \to 0}f(x)=0$ and $\lim_{x \to 0}g(x)=\infty$):-$$\begin{align}\lim_{x \to 0}(1+f(x))^{g(x)}&=\lim_{x \to 0}e^{g(x)\ln[1+f(x)]}\\&=e^{\lim_{x \to 0}g(x)[1+f(x)-1]}\\&=e^{\lim_{x \to 0}g(x)f(x)}\end{align}$$Here, I have used the fact that $\lim_{h(x) \to 1}\ln(h(x))=h(x)-1$. In most of these questions, it is easy to find $\lim_{x \to 0}g(x)f(x)$ which is of the form $\infty*0$. For your question $f(x)=-x$ and $g(x)=\frac 1x$ (In your limit $n \to \infty$ and in my limit $x\to0$, hence the difference). So $\lim_{x\to0}g(x)f(x)=-1$. Therefore your limit is $e^{-1}=\frac 1e$.

Answer 5

If one wants to use the squeeze theorem, then we can proceed as follows.

First, We define

$$e=\lim_{n\to \infty}\left(1+\frac1n\right)^n \tag 1$$

Then, we note that

$$\left(1-\frac1n\right)\left(1+\frac1n\right)=\left(1-\frac{1}{n^2}\right)<1\implies \left(1-\frac1n\right)^n <\frac{1}{\left(1+\frac1n\right)^n } \tag 2$$

Also, using Bernoulli's Theorem we have

$$\left(1-\frac{1}{n^2}\right)^n\ge 1-\frac1n\implies \left(1-\frac1n\right)^n \ge \frac{1-\frac1n}{\left(1+\frac1n\right)^n} \tag 3$$

Putting $(2)$ and $(3)$ together reveals that

$$\frac{1-\frac1n}{\left(1+\frac1n\right)^n} \le \left(1-\frac1n\right)^n \le \frac{1}{\left(1+\frac1n\right)^n }$$

whence taking the limit as $n\to \infty$, using $(1)$ along with the squeeze theorem provides the anticipated result

$$\lim_{n\to \infty}\left(1-\frac1n\right)^n=e^{-1}$$

Answer 6

Another way considering $$A=\left( 1-\frac{1}{n}\right)^{n}$$ Using logarithms $$\log(A)=n\log\left( 1-\frac{1}{n}\right)$$ Now, using Taylor expansion for small values of $x$ $$\log(1-x)=-x-\frac{x^2}{2}+O\left(x^3\right)$$ replacing $x$ by $\frac 1n$ $$\log(A)=n\left(-\frac 1 n -\frac{1}{2n^2}+\cdots\right)=-1-\frac{1}{2n}+\cdots$$

I am sure that you can take from here.