hallo,
does any one knows how to prove that lim(1+1/n)^n=e when n goes to zero.
thanks
omri
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, $\displaystyle L = 1.$
$\displaystyle \lim_{n\to\infty}\left ( 1 + \frac{1}{n} \right )^n=e$
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.
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 $\displaystyle \ln\varphi=1\,\therefore\,\varphi=e.\quad\blacksqu are$
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.
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.
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.