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

Question

How would one compute the following limit?

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

I know

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

but right there is a minus keeping that limit from being used.

Another problem I am questioning is finding the limit of

$$\frac{n!}{2n}$$

Of course, $\frac{x^n}{n!}$ has zero as a limit but here it is the opposite.

$$\left(1 - \frac1n\right)^n = \left(1-\frac1n\right)\cdot \frac{1}{\left(1 + \frac{1}{n-1}\right)^{n-1}}$$ — Daniel Fischer, Oct 07 '13 at 19:46
Hmm how did you get that $=(1-\frac{1}{n}$ $\frac{1}{1+1/n-1}^{n-1})$ — Fernando Martinez, Oct 07 '13 at 19:54
yes when you take the limit n approach infinity it does equal 1 — Fernando Martinez, Oct 07 '13 at 19:57
Yet another way $\lim_{n \to \infty} (1-\frac1n)^n = \lim_{n \to -\infty} ((1+\frac1n)^n)^{-1}=e^{-1}$ — chubakueno, Oct 07 '13 at 20:21
Possible duplicate of About $\lim \left(1+\frac {x}{n}\right)^n$ — Xander Henderson, Mar 06 '19 at 13:17

score 15 · Accepted Answer · answered Oct 07 '13 at 20:04

15

You can obtain the limit of

$$\left(1-\frac1n\right)^n$$

easily from the one you know, $\left(1 + \frac1n\right)^n \to e$, by noting

$$1 - \frac1n = \frac{n-1}{n} = \frac{1}{\left(\frac{n}{n-1}\right)} = \frac{1}{\left(1 + \frac{1}{n-1}\right)}.$$

Then you can write

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

where the first factor obviously converges to $1$, and the second one converges to $\frac{1}{e}$ by what you already know.

answered Oct 07 '13 at 20:04

Daniel Fischer

206,697

yes that make sense one thing is the part when you go from $\frac{1}{n/(n-1)}$ to $\frac{1}{1+(1/n-1)}$ how did you add the plus one – Fernando Martinez Oct 07 '13 at 20:10
@FernandoMartinez : $n/(n-1) = (n-1+1)/(n-1) = 1 + 1/(n-1)$. – Théophile Oct 07 '13 at 20:24
I get the $(n/(n-1)=(n-1+1)/(n-1)$ part but the next step confounds me – Fernando Martinez Oct 07 '13 at 20:29
2

Which next step? $$\frac{(n-1)+1}{n-1} = \frac{(n-1)}{n-1} + \frac{1}{n-1} = 1 + \frac{1}{n-1},?$$ – Daniel Fischer Oct 07 '13 at 20:30
yes senor fischer that part – Fernando Martinez Oct 07 '13 at 20:31
I've just edited the step in, does that clear it up? – Daniel Fischer Oct 07 '13 at 20:32
oh yes it does I thought you added n-1+1 giving you n but I see now my error i am gratefull – Fernando Martinez Oct 07 '13 at 20:34

score 3 · Answer 2 · answered Oct 07 '13 at 19:56

3

For another one use the definition of $n!$. In fact:

$$\frac{n!}{2n}>>n$$

answered Oct 07 '13 at 19:56

Mikasa

67,374

$\ddot\smile ;+1$ – amWhy Oct 07 '13 at 23:30
Beauty is in the simplicity :)+1 – Oct 08 '13 at 18:51
@SamiBenRomdhane: Thanks for saying so ;-) – Mikasa Oct 11 '13 at 17:19

daulomb · Answer 3 · 2013-10-07T20:09:46.757

2

The result of the limit is $e^{-1}$ not $e^n$. Let $y= (1-\frac{1}{n})^n$, or $\ln y= n\ln (1-\frac{1}{n})$. Since $\displaystyle\lim_{n\rightarrow\infty}n\ln (1-\frac{1}{n})=\lim_{n\rightarrow\infty}\frac{\ln (1-\frac{1}{n})}{1/n}=-1$ by L'Hospital's rule, so $y=\frac{1}{e}$. The other limit is infinity: $\lim\frac{n!}{x^n}=\lim\frac{1}{\frac{x^n}{n!}}=\infty$. In the same manner since $\lim\frac{2n}{n!}=0$, we have $\lim\frac{n!}{2n}=\infty$.

edited Oct 07 '13 at 20:09

answered Oct 07 '13 at 19:48

daulomb

3,955

this must be a comment not an answer . – what'sup Oct 07 '13 at 19:49

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

3 Answers3

Linked