1. the limit of (1+1/n)^n=e

hallo,

does any one knows how to prove that lim(1+1/n)^n=e when n goes to zero.

thanks

omri

2. Originally Posted by omrimalek
hallo,

does any one knows how to prove that lim(1+1/n)^n=e when n goes to zero.

thanks

omri
I assume that you made a typo because $\displaystyle \lim_{n \mapsto 0}\left(1+\frac1n\right)^n = 1$

3. Hi

$\displaystyle L=\lim_{n \to 0} \left(1+\frac 1n\right)^n$

$\displaystyle \ln(L)=\lim_{n \to 0} n \cdot \ln \left(1+\frac 1n\right)$

But when $\displaystyle n \to 0 ~,~ \frac 1n \gg 1$

Therefore $\displaystyle \ln(L)=\lim_{n \to 0} n \cdot \ln \left(\frac 1n\right)$

Substituting $\displaystyle u=\frac 1n$, we get :

$\displaystyle \ln(L)=\lim_{u \to \infty} \frac 1u \cdot \ln(u)=0$.

Therefore, $\displaystyle L = 1.$

4. Originally Posted by earboth
I assume that you made a typo because $\displaystyle \lim_{n \mapsto 0}\left(1+\frac1n\right)^n = 1$
May be he means as $\displaystyle n \to \infty$??

Which can be done by taking logs then showing the limit is $\displaystyle 1$ by any number of methods including L'Hopitals rule.

RonL

5. sorry i made a mistake...

i ment when n goes to infinity...

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

This is the definition of number e. So we can't prove that this is e, but we can show that this definition is consistent with other rules and definitions.

7. Originally Posted by wingless
$\displaystyle \lim_{n\to\infty}\left ( 1 + \frac{1}{n} \right )^n=e$

This is the definition of number e. So we can't prove that this is e, but we can show that this definition is consistent with other rules and definitions.
It can be proved but that requires assuming some other definition which would otherwise be proved from this definition .....

What can be done without assuming any definitions is proving that the limit exists and that it lies between 2 and 3.

8. thank you all!

9. Let $\displaystyle \varphi = \lim_{x \to \infty } \left( {1 + \frac{1} {x}} \right)^x .$ (Assuming the limit does exist.)

Since the logarithm is continuous on its domain, we can interchange the function and taking limits.

$\displaystyle \ln \varphi = \ln \Bigg[ {\lim_{x \to \infty } \left( {1 + \frac{1} {x}} \right)^x } \Bigg] = \lim_{x \to \infty } x\ln \left( {1 + \frac{1} {x}} \right).$

Make the substitution $\displaystyle u=\frac1x,$

$\displaystyle \ln \varphi =\lim_{u\to0}\frac1u\ln(1+u).$ Since $\displaystyle \ln (1 + u) = \int_1^{1 + u} {\frac{1} {\tau }\,d\tau } ,\,1 \le\tau\le 1 + u,$

$\displaystyle \frac{1} {{1 + u}} \le \frac{1} {\tau } \le 1\,\therefore \,\frac{u} {{1 + u}} \le \int_1^{1 + u} {\frac{1} {\tau}\,d\tau} \le u.$ So $\displaystyle \frac1{1+u}\le\frac1u\ln(1+u)\le1.$

Take the limit when $\displaystyle u\to0,$ then by the Squeeze Theorem we can conclude that $\displaystyle \lim_{u\to0}\frac1u\ln(1+u)=1.$

Finally $\displaystyle \ln\varphi=1\,\therefore\,\varphi=e.\quad\blacksqu are$

10. Originally Posted by Moo
Hi
You just have to be careful about one thing over here. You assumed the limit exists. How do you know the limit exists? Note, this is why Krizalid wrote "assuming it exists". To prove this we can note the function $\displaystyle f0,\infty)\mapsto \mathbb{R}$ defined as $\displaystyle f(x) = (1+\tfrac{1}{x})^x$ is increasing since $\displaystyle f' >0$. Therefore, the sequence $\displaystyle x_n = (1+\tfrac{1}{n})^n$ is an increasing sequence. Now show that $\displaystyle \{ x_n\}$ is bounded. So we have a increasing bounded sequence and therefore we have a limit.

Originally Posted by wingless
$\displaystyle \lim_{n\to\infty}\left ( 1 + \frac{1}{n} \right )^n=e$

This is the definition of number e. So we can't prove that this is e, but we can show that this definition is consistent with other rules and definitions.
That is not how I like to define $\displaystyle e$. I like to define $\displaystyle \log x = \smallint_1^x \tfrac{d\mu}{\mu}$. And we can define $\displaystyle e$ to be the (unique) number such that $\displaystyle \log (e) = 1$. If we define $\displaystyle e$ this way then it would follow that $\displaystyle (1+\tfrac{1}{n})^n\to e$. But whatever, it depends on your style of defining logarithmic functions. I just find that the approach I use it the cleanest and smoothest.

11. Originally Posted by wingless
$\displaystyle \lim_{n\to\infty}\left ( 1 + \frac{1}{n} \right )^n=e$

This is the definition of number e. So we can't prove that this is e, but we can show that this definition is consistent with other rules and definitions.
It is not $\displaystyle the$ definition but $\displaystyle a$ definition, how one proves it depends on what one is supposed to know, and how it has been defined. It is quite common for it to be defined as the base of natural logarithms.

RonL

12. Originally Posted by CaptainBlack
It is not $\displaystyle the$ definition but $\displaystyle a$ definition, how one proves it depends on what one is supposed to know, and how it has been defined. It is quite common for it to be defined as the base of natural logarithms.

RonL
Thanks to you both, TPH and CaptainBlack. I know that there are tons of definitions for e, but this definition is the oldest one so it seemed me nonsense to prove it.

13. I will present this proof for what it's worth. Though, the posters proof builds

from this. This proof builds on the differentiability of ln(x). To be more exact

the

derivative of ln(x) at x=1.

Using the definition of a derivative, since 1/x=1, we get:

$\displaystyle 1=\lim_{h\to 0}\frac{ln(1+h)-ln(1)}{h}$

$\displaystyle =\lim_{h\to 0}\frac{ln(1+h)}{h}$

$\displaystyle =\lim_{h\to 0} ln(1+h)^{\frac{1}{h}}$

Therefore, it follows that:

$\displaystyle e=e^{\lim_{h\to 0} ln(1+h)^{\frac{1}{h}}}$

from the continuity of $\displaystyle e^{x}$ can be written this way:

$\displaystyle e=\lim_{h\to 0}e^{ln(1+h)^{\frac{1}{h}}}=\lim_{h\to 0}(1+h)^{\frac{1}{h}}$

Now, we recognize this limit. The others play off it.

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

merely let $\displaystyle t=\frac{1}{x}$.

This changes the limit to $\displaystyle x\to 0$ and we have said limit.