Limit of $\left(1+\frac{1}{2n}\right)^n$

Question

$$\lim\limits_{n\to\infty}(1+\frac{1}{2n})^n=\sqrt{e}$$

But I could not reason why this is, I know that I can rewrite it as

$$\lim\limits_{n\to\infty}(1+\frac{\frac{1}{2}}{n})^n=\sqrt{e}$$

But I can't find any real calculation where it is actually calculated, and at least for me, it's not totally obvious.

Or in other words its not obvious why

$$\lim\limits_{n\to\infty}(1+\frac{x}{n})^n=e^x$$

Maybe there is a link which I could not find, where the relation is explained. Many thanks.

The definition/identity $e^x=\lim_{n\to\infty}(1+x/n)^n$ is very well-known. See e.g. this answer. — Parcly Taxel, Mar 09 '19 at 13:34
Do you know that $\lim\limits_{n\to\infty}(1+\frac{1}{n})^n={e}$? — cqfd, Mar 09 '19 at 13:36
$\lim\limits_{n\to\infty}(1+\frac{1}{2n})^n$ now take $m=2n$, and obviously $\lim\limits_{m\to\infty} as \lim\limits_{n\to\infty}$, so you get $\lim\limits_{m\to\infty}(1+\frac{1}{m})^{m/2}$. I hope the rest of the logic is easy. — Aven Desta, Mar 09 '19 at 13:38
@ThomasShelby yes, i actually just understood the problem :) many thanks for your answer — Ang, Mar 09 '19 at 13:51

score 3 · Answer 1 · answered Mar 09 '19 at 13:38

3

Write your term in the form $$\left(\left(1+\frac{1}{2n}\right)^{2n}\right)^{1/2}$$

answered Mar 09 '19 at 13:38

Dr. Sonnhard Graubner

95,283

Dr. Mathva · Answer 2 · 2019-03-09T13:55:51.800

3

As suggested in the comments, consider the well-known identity

$$\lim_{n\to\infty}\bigg(1+\frac{x}{n}\bigg)^n=e^x\tag1$$ Now let $x=\frac{1}{2}$ in order to obtain

$$\lim_{n\to\infty}{\bigg(1+\frac{1}{2n}}\bigg)^n=e^{\frac{1}{2}}=\sqrt e$$

Now, if you want to understand the identity (1), you might want to check this

edited Mar 09 '19 at 13:55

answered Mar 09 '19 at 13:38

Dr. Mathva

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Thanks i was looking for a way to understand the identity :) many thanks for your help! – Ang Mar 09 '19 at 13:53
@Ang I'm glad I could help you ;) – Dr. Mathva Mar 10 '19 at 17:02

score 3 · Accepted Answer · answered Mar 09 '19 at 13:44

3

Hint:

Compute the limit of the log: $$\ln\biggl(1+\frac1{2n}\biggr)^{\negthickspace n}=n\ln\biggl(1+\frac1{2n}\biggr)=\frac{\ln\biggl(1+\frac1{2n}\biggr)}{\frac1n}\sim_\infty\frac{\frac1{2\not n}}{\frac1{\not n}}=\frac12.$$

answered Mar 09 '19 at 13:44

Bernard

175,478

Michael Rybkin · Answer 4 · 2019-03-09T13:55:43.237

2

Are you familiar with this very famous limit:

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

You can find a very good proof of why it's equal to $e^r$ here.

edited Mar 09 '19 at 13:55

answered Mar 09 '19 at 13:37

Michael Rybkin

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Yeah i was looking for the proof :) many thanks for your answer!! – Ang Mar 09 '19 at 13:53

Limit of $\left(1+\frac{1}{2n}\right)^n$

4 Answers4