# Euler's number

• Apr 23rd 2008, 10:19 PM
a.a
Euler's number
The number is defined as the limit of (1+1/n)^n as n approched infinity.
My question is why is that limit the number e. If you evaluate it useing the subsistutaion method, you get the limit is 1.

E is also defined as the limit of (1+n)^(1/n) as n approaches 0. Using the substitution method isn't this limit also 1?
• Apr 23rd 2008, 11:15 PM
Isomorphism
Quote:

Originally Posted by a.a
My question is why is that limit the number e. If you evaluate it useing the subsistutaion method, you get the limit is 1.

Could you please shows that substitution method and show us how the limit evaluates to 1?

On the other hand $\displaystyle \left(1 + \frac1{n}\right)^n = 1 + {n \choose 1} \frac1{n} + {n \choose 2} \frac1{n^2} + ...$.

Thus $\displaystyle \forall n \in \mathbb{N}\, , \, \left(1 + \frac1{n} \right)^n > 1 + {n \choose 1} \frac1{n}$

Which means $\displaystyle \forall n \in \mathbb{N}\, ,\, \left(1 + \frac1{n}\right)^n > 2$

This means for all natural numbers the expression is always greater than 2... How can the limit be 1 then?
• Apr 23rd 2008, 11:20 PM
a.a
Quote:

Originally Posted by Isomorphism
Could you please shows that substitution method and show us how the limit evaluates to 1?

lim (1+ 1/n)^n as n approaches infinity
= (1+ 0)^enfinity
= 1

lim (1+ n) ^(1/n) as n approaches 0
= (1+0)^enfinity
= 1
• Apr 23rd 2008, 11:22 PM
o_O
$\displaystyle 1^{\infty}$ is an indeterminate form just like $\displaystyle 0^{0}$ is where the limit depends on the functions you have in place of 1, 0, or infinity. $\displaystyle 1^{\text{anything}}$ isn't always 1. For example:
$\displaystyle \lim_{x \to 0} \cos (x)^{\frac{1}{x^{2}}} = \left[1^{\infty}\right] = e^{-\frac{1}{2}}$

How?
$\displaystyle \lim_{x \to 0} \cos (x)^{\frac{1}{x^{2}}} = e^{\lim_{x \to 0} \ln \cos x^{1/x^{2}}}$

Looking at the exponent:
$\displaystyle \lim_{x \to 0} \ln \cos x^{1/x^{2}} = \lim_{x \to 0} \frac{\ln \cos x}{x^{2}} = \left[\frac{0}{0}\right]$
$\displaystyle = \lim_{x \to 0} \frac{-\tan x}{2x} = \left[\frac{0}{0}\right]$
$\displaystyle = \lim_{x \to 0} \frac{-sec^{2} x}{2} = - \frac{1}{2}$

Going back to the original limit, we see that:
$\displaystyle \lim_{x \to 0} \cos (x)^{\frac{1}{x^{2}}} = e^{-\frac{1}{2}} \neq {\color{red} 1}$
• Apr 23rd 2008, 11:24 PM
Isomorphism
Quote:

Originally Posted by a.a
lim (1+ 1/n)^n as n approaches infinity
= (1+ 0)^enfinity
= 1

lim (1+ n) ^(1/n) as n approaches 0
= (1+0)^enfinity
= 1

Speaking loosely, $\displaystyle 1^{\infty} \neq 1$.That is your mistake

Look at my old post again. I have proved that (1+ 1/n)^n is always greater than 2.