1. ## The "e" limit

$e=\lim_{x\to\infty}\bigg(1+\frac1x\bigg)^x$

What's the proof of this?

2. Well, this could easily be a definition

But just using Taylor series for log, we get:

$\lim_{x\rightarrow\infty}(1+\frac{1}{x})^x=\lim_{x \rightarrow\infty}{\rm e}^{x\log(1+\frac{1}{x})}=\lim_{x\rightarrow\infty }{\rm e}^{x\left(\frac{1}{x}-\frac{1}{2x^2}+\frac{1}{3x^3}+...\right)}=\lim_{x\ rightarrow\infty}{\rm e}^{1-\frac{1}{2x}+\frac{1}{3x^2}+...}$

and by continuity of the exponential function, this is equal to

${\rm e}^{\lim_{x\rightarrow\infty}\left(1-\frac{1}{2x}+\frac{1}{3x^2}+...\right)}={\rm e}$

3. Ohhh... and without Taylor Series?

4. Originally Posted by Krizalid
... and without Taylor Series?
Can you prove this inequality $0 < a < b\quad \Rightarrow \quad \frac{1}{b} < \frac{{\ln (b) - \ln (a)}}{{b - a}} < \frac{1}{a}$?

Let $a = 1\quad \& \quad b = 1 + \frac{1}{x}$ using the inequality we get
$\frac{1}{{x + 1}} < \ln \left( {1 + \frac{1}{x}} \right) < \frac{1}{x}$.

Using the property of exponents:
$e^{\left( {\frac{1}{{x + 1}}} \right)} < \left( {1 + \frac{1}{x}} \right) < e^{\left( {\frac{1}{x}} \right)} \quad \Rightarrow e^{\left( {\frac{x}{{x + 1}}} \right)} < \left( {1 + \frac{1}{x}} \right)^x < e\quad$.

Now the limit is apparent.

5. Originally Posted by Krizalid
$e=\lim_{x\to\infty}\bigg(1+\frac1x\bigg)^x$

What's the proof of this?
One thing that you have to think about when faced with this question is:

What definition of e do you have to work with?

RonL

6. Well... the base of natural logarithms.

Can you show a proof of what I'm askin'?

7. Originally Posted by Krizalid
Well... the base of natural logarithms.

Can you show a proof of what I'm askin'?
Well I would consider:

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

and show that this satisfies the initial value problem:

$
\frac{df}{dx}=f(x),\ f(0)=1
$

RonL

8. Originally Posted by CaptainBlack
Well I would consider:

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

and show that this satisfies the initial value problem:

$
\frac{df}{dx}=f(x),\ f(0)=1
$
Aha... how'd be the proof?

9. Originally Posted by CaptainBlack
Well I would consider:

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

and show that this satisfies the initial value problem:

$
\frac{df}{dx}=f(x),\ f(0)=1
$

RonL
Originally Posted by Krizalid
Aha... how'd be the proof?
Put:

$
f_n(x)= \left( 1+\frac{x}{n}\right)^n
$

then:

$
\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:

$
\frac{df}{dx}=\lim_{n \to \infty} \frac{df_n}{dx}=$
$\frac{\lim_{n \to \infty}f_n(x)}{\lim_{n \to \infty}(1+x/n)}=f(x)
$

Now as $f_n(0)=1 \ \forall n,\ \ f(0)=1$.

So we have $f(x)$ is the solution of the initial value problem:

$f'(x)=f(x), \ \ f(0)=1$

which is all we need.

RonL

10. What definition of e do you have to work with?
Well... the base of natural logarithms.
If you ever write a book, don't use this definition, it's going to be hell to derive basic properties

11. Krizalid, the stadard way $e^x$ is defined is first by defining,
$\ln x = \int_1^x \frac{dt}{t} \mbox{ for }x>0$.
And let $e^x = \ln ^{-1} x$.

12. There's actually many ways to define e.

The one I like most is $e=\lim(1+1/n)^n$, which comes natural historically (from medieval banking, that is )

Another relatively easy way is by the (all familiar) series $e=\sum_n 1/n!$, but this is troublesome when it comes to deriving the basic properties of $e^x$.

The most delicate way, is by describing the function $\exp:\mathbb{R}\rightarrow\mathbb{R}$ to be the unique solution of the functional equation $\exp(x+y)=\exp(x)\exp(y)$ to satisfy $\exp(0)=1$. Then $e=\exp(1)$.

Ok, I know you are bored by now, so I stop here ()

13. Originally Posted by Rebesques
There's actually many ways to define e.
But they are not as easy to use for analytical arguments as the $\ln x$ one. If you were writing an analysis book I am sure you will use $\ln x$ approach because it is by far the simplest to derive all the identities and properties for $e^x$.

The most delicate way, is by describing the function $\exp:\mathbb{R}\rightarrow\mathbb{R}$ to be the unique solution of the functional equation $\exp(x+y)=\exp(x)\exp(y)$ to satisfy $\exp(0)=1$. Then $e=\exp(1)$.
What about $\exp (x) = 1$?

14. But they are not as easy to use for analytical arguments as the...
I' ll disagree here, the lim(1+1/n)^n works fine to that extent.

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 post

15. Here's a proof which involves integrals (But it's just a little integral.)

Let

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

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

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

Make the substitution $u=\frac1x,$

$\ln \varphi =\lim_{u\to0}\frac1u\ln(1+u).$

Since $\ln (1 + u) = \int_1^{1 + u} {\frac{1}
{\tau }\,d\tau } ,\,1 \le\tau\le 1 + u,$

$\frac{1}
{{1 + u}} \le \frac{1}
{\tau } \le 1\,\therefore \,\frac{u}
{{1 + u}} \le \int_1^{1 + u} {\frac{1}
{\tau}\,d\tau} \le u.$

So $\frac1{1+u}\le\frac1u\ln(1+u)\le1.$

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

Finally $\ln\varphi=1\,\therefore\,\varphi=e.$