proof of that the definition limit of "e" exists... please help!

**rargh** · October 1st 2010, 10:01 AM

Hi,

I'm reviewing some basic calculus concepts, and I have the following questions that will probably be very easy for you guys to answer, although I'm stuck since a long time on them.

Given the usual limit that is used in the definition of "e": lim (n->infinity) (1+1/n), how do we prove that the limit exists and it is finite?

Can we use the binomial theorem expansion for a finite n, and prove that the coefficients tend to be 1/k! for n that goes to infinity, so we get the Taylor expansion? I'd like to see a rigorous proof of this derivation of e's expression as a power series, we shouldn't forget to prove that the coefficients of the binomial expansion don't diverge (in fact, i guess they should converge) as n goes to infinity...

Or without expanding it, can we prove the following statements:

let a(n)=(1+1/n)^n

1-prove that a(n) is strictly increasing
2-prove that a(n) is limited above by a finite positive number L (for instance, we could try to prove it for L=3 or L=4): a(n)<L for all n.

Final questions:

A-if we manage to prove the two statements above, then did we successfully prove that the limit exists?
B-is there a better way than this to prove that the limit exists?

Thanks! And please excuse me for the length of the post...

**Plato** · October 1st 2010, 10:25 AM

Using the notation $e^x=\exp(x)$ we show it.
It is easy to use Napier’s inequality to show that $\dfrac{1}{n+1}\le\ln\left(1+\frac{1}{n}\right)\le\ dfrac{1}{n}.$

Raise e to those powers $\exp\left(\frac{1}{n+1}\right)\le\left(1+\frac{1}{ n}\right)\le\exp\left(\frac{1}{n}\right).$ .

Raise each of those to the nth power
$\exp\left(\frac{n}{n+1}\right)\le\left(1+\frac{1}{ n}\right)^n\le e.$ .

It is squeezed between e & e.

**rargh** · October 1st 2010, 10:54 AM

Thank you for answering so soon, the proof is very simple, but how do you define ln(x) without proving the limit of e first? I mean, wouldn't it be circular logic?

We should find a way to prove it without using the natural logarithm...

Alternatively, we could define ln(x) as the integral of 1/t from 1 to x, and then napier's inequality should be simple to infer, but we need to use a integral definition...

there should be an elementary method to prove it...

**Plato** · October 1st 2010, 11:10 AM

There are so many different ways to do this.

Notice that for $n\ge 2$ we have $\[ \left( {1 - \frac{1}{{n^2 }}} \right)^n \geqslant \left( {1 - \frac{1} {n}} \right)\; \Rightarrow \;\left( {1 + \frac{1}{n}} \right)^n \geqslant \left( {1 + \frac{1}{{n - 1}}} \right)^{n - 1}$ .

So the sequence $\left( {1 + \frac{1}{n}} \right)^n$ is increasing and bounded above. That means it converges.

Call that limit $e$ .

EDIT: If you need to show it is bounded.
$\left( {1 + \frac{1} {n}} \right)^n = \sum\limits_{k = 0}^n {C(n,k)\left( {\frac{1} {n}} \right)^k } < 1 + \left( {1 + \frac{1} {2} + \cdots + \frac{1} {{2^{n - 1} }}} \right) < 3$

**rargh** · October 1st 2010, 11:40 AM

Ok thank you very much, I understood how you deduced the facts from those inequalities, I will work to prove them.

You have been very kind, thanks again.

**emakarov** · October 1st 2010, 02:17 PM

Here is a couple of links from PlanetMath.

**rargh** · October 1st 2010, 10:45 PM

Great, thanks again!

**Prove It** · October 2nd 2010, 12:21 AM

By definition, $e$ is the base of the exponential function that is its own derivative.

Therefore $\frac{d}{dx}(e^x) = e^x$ .

If we wish to evaluate $e$ , we need to use this definition.

$\frac{d}{dx}(e^x) = \lim_{h \to 0}\frac{e^{x + h} - e^x}{h}$

$= \lim_{h \to 0}\frac{e^xe^h - e^x}{h}$

$= \lim_{h \to 0}\frac{e^x(e^h - 1)}{h}$

$= e^x\lim_{h \to 0}\frac{e^h - 1}{h}$ .

Since we know by definition that $\frac{d}{dx}(e^x) = e^x$ , that means

$e^x\lim_{h \to 0}\frac{e^h - 1}{h} = e^x$

$\lim_{h \to 0}\frac{e^h - 1}{h}= 1$

$\lim_{h \to 0}(e^h - 1) = \lim_{h \to 0}h$

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

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

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

Now if we let $h = \frac{1}{n}$ , then as $h \to 0, n \to \infty$ .

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

**HallsofIvy** · October 2nd 2010, 02:40 AM

Another useful way to define ln(x) is simply:
$ln(x)= \int_1^x \frac{1}{t}dt$

Since $\frac{1}{t}$ is defined for all non-zero t, this defines ln(x) for all positive x. It is clearly continuous and differentiable for all x. By the fundamental theorem of Calculus, its derivative is 1/x. Since that is positive for all positive x, ln(x) is an increasing function and so a one-to-one function.

It is not difficulty to prove, by substitutions in the integral, that
1) For any positive x, $ln(1/x)= - ln(x)$ .
2) For any positive x and y, $ln(xy)= ln(x)+ ln(y)$ .
3) For any positive x and y any real number, $ln(x^y)= y ln(x)$ .

For example, for the last one, $ln(x^y)$ is, by this definition, $\int_1^{x^y}\frac{1}{t}dt$ .

If $y\ne 0$ , define $u= t^{1/y}$ so that $t= u^y$ , $du= y u^{y- 1}du$ so that $\frac{1}{t}dt= \frac{1}{u^y}(y u^{y-1}du)= y \frac{1}{u}du$ . When t= 1, $u= 1^{1/y}= 1$ and when $t= x^y$ , $u= (x^y)^{1/y}= x$ so the integral becomes
$ln(x^y)= \int_1^x y \frac{1}{u}du= y\int_1^x \frac{1}{u}du= y ln(x)$

(If y= 0 then $x^y= x^0= 1$ and $ln(x^y)= ln(1)= \int_1^1 \frac{1}{t}dt= 0= 0 ln(x)= y ln(x)$ still.)

It follows then that ln(x) is a one-to-one function from the positive real numbers onto the real numbers and so has an inverse function, exp(x), from the real numbers to the positive real numbers.

Finally, if y= exp(x), then x= ln(y). If x is not 0, divide both sides by x to get $1= \frac{1}{x}ln(y)= ln(y^{1/x})$ . Going back to the exponential form, $y^{1/x}= exp(1)$ so $y= (exp(1))^x$ .
(If x= 0, then y= 1 because $ln(1)= \int_1^1 \frac{1}{t}dt= 0$ . So we still have $exp(x)= exp(0)= 1= (exp(1))^0$ .

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