$\displaystyle e=\lim_{x\to\infty}\bigg(1+\frac1x\bigg)^x$
What's the proof of this?
Well, this could easily be a definition
But just using Taylor series for log, we get:
$\displaystyle \lim_{x\rightarrow\infty}(1+\frac{1}{x})^x=\lim_{x \rightarrow\infty}{\rm e}^{x\log(1+\frac{1}{x})}=\lim_{x\rightarrow\infty }{\rm e}^{x\left(\frac{1}{x}-\frac{1}{2x^2}+\frac{1}{3x^3}+...\right)}=\lim_{x\ rightarrow\infty}{\rm e}^{1-\frac{1}{2x}+\frac{1}{3x^2}+...}$
and by continuity of the exponential function, this is equal to
$\displaystyle {\rm e}^{\lim_{x\rightarrow\infty}\left(1-\frac{1}{2x}+\frac{1}{3x^2}+...\right)}={\rm e}$
Can you prove this inequality $\displaystyle 0 < a < b\quad \Rightarrow \quad \frac{1}{b} < \frac{{\ln (b) - \ln (a)}}{{b - a}} < \frac{1}{a}$?
Let $\displaystyle a = 1\quad \& \quad b = 1 + \frac{1}{x}$ using the inequality we get
$\displaystyle \frac{1}{{x + 1}} < \ln \left( {1 + \frac{1}{x}} \right) < \frac{1}{x}$.
Using the property of exponents:
$\displaystyle e^{\left( {\frac{1}{{x + 1}}} \right)} < \left( {1 + \frac{1}{x}} \right) < e^{\left( {\frac{1}{x}} \right)} \quad \Rightarrow e^{\left( {\frac{x}{{x + 1}}} \right)} < \left( {1 + \frac{1}{x}} \right)^x < e\quad $.
Now the limit is apparent.
Put:
$\displaystyle
f_n(x)= \left( 1+\frac{x}{n}\right)^n
$
then:
$\displaystyle
\frac{ df_n}{d x}=n (1+x/n)^{n-1} \frac{1}{n}=f_n(x)/(1+x/n)
$
Now I will assume that you can interchange the two limits involved to get:
$\displaystyle
\frac{df}{dx}=\lim_{n \to \infty} \frac{df_n}{dx}=$$\displaystyle \frac{\lim_{n \to \infty}f_n(x)}{\lim_{n \to \infty}(1+x/n)}=f(x)
$
Now as $\displaystyle f_n(0)=1 \ \forall n,\ \ f(0)=1$.
So we have $\displaystyle f(x)$ is the solution of the initial value problem:
$\displaystyle f'(x)=f(x), \ \ f(0)=1$
which is all we need.
RonL
There's actually many ways to define e.
The one I like most is $\displaystyle e=\lim(1+1/n)^n$, which comes natural historically (from medieval banking, that is )
Another relatively easy way is by the (all familiar) series $\displaystyle e=\sum_n 1/n!$, but this is troublesome when it comes to deriving the basic properties of $\displaystyle e^x$.
The most delicate way, is by describing the function $\displaystyle \exp:\mathbb{R}\rightarrow\mathbb{R}$ to be the unique solution of the functional equation $\displaystyle \exp(x+y)=\exp(x)\exp(y)$ to satisfy $\displaystyle \exp(0)=1$. Then $\displaystyle e=\exp(1)$.
Ok, I know you are bored by now, so I stop here ()
But they are not as easy to use for analytical arguments as the $\displaystyle \ln x$ one. If you were writing an analysis book I am sure you will use $\displaystyle \ln x$ approach because it is by far the simplest to derive all the identities and properties for $\displaystyle e^x$.
What about $\displaystyle \exp (x) = 1$?The most delicate way, is by describing the function $\displaystyle \exp:\mathbb{R}\rightarrow\mathbb{R}$ to be the unique solution of the functional equation $\displaystyle \exp(x+y)=\exp(x)\exp(y)$ to satisfy $\displaystyle \exp(0)=1$. Then $\displaystyle e=\exp(1)$.
I' ll disagree here, the lim(1+1/n)^n works fine to that extent.But they are not as easy to use for analytical arguments as the...
You just verified that there were three conditions altogether in that definiton, I just thought hard but couldn't remember if there was another one when making that postWhat about... ?
Here's a proof which involves integrals (But it's just a little integral.)
Let
$\displaystyle \varphi = \lim_{x \to \infty } \left( {1 + \frac{1}
{x}} \right)^x .$
Since the logarithm is continuous on its domain, we can interchange the function and taking limits.
$\displaystyle \ln \varphi = \ln \left[ {\lim_{x \to \infty } \left( {1 + \frac{1}
{x}} \right)^x } \right] = \lim_{x \to \infty } x\ln \left( {1 + \frac{1}
{x}} \right).$
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 $\displaystyle \ln\varphi=1\,\therefore\,\varphi=e.$