# Proof for exponentials

1. Mar 14, 2006

### island-boy

hello, I need the proof to show that:
$$e^x = \lim_{n\rightarrow\infty} (1+x/n)^n$$

Here's what I was able to come up with so far:
since the derivative of $$e^x$$ is also $$e^x$$,

then let f(x) = $$e^x$$
thus:
D(f(x)) = $$\lim_{n\rightarrow\infty} \frac{f(x+h) - f(x)}{h} = \lim_{n\rightarrow\infty} \frac{e^{x+h} - e^x}{h} = e^x\lim_{n\rightarrow\infty} \frac{e^h - 1}{h}$$

so for the derivative of $$e^x$$ to equal itself,
$$\lim_{n\rightarrow\infty} \frac{e^h - 1}{h} = 1$$

so for small values of h, we can write:
$$e^h - 1 = h$$
and so
$$e = (1+h)^{1/h}$$

Replacing h by 1/n, we get:
$$e = (1 + 1/n)^n$$
As n gets larger and approaches infinity, we get:
$$e = \lim_{n\rightarrow\infty} (1+1/n)^n$$

so, how do I get:
$$e^x = \lim_{n\rightarrow\infty} (1+x/n)^n$$

Also, is it true that:
$$(1+x/n)^n \leq e^x$$ and
$$(1-x/n)^n \leq e^{-x}$$ for every natural n and every x element of X?
How can I prove this?

thanks!

Last edited: Mar 14, 2006
2. Mar 14, 2006

### Fermat

Have you tried the Binomial theorem ?

3. Mar 14, 2006

### Hurkyl

Staff Emeritus
I would've taken the expression for e and raised both sides to the x power.

You really ought to do this more rigorously...

P.S. many of the limits in your limits are wrong.

4. Mar 14, 2006

### arildno

As a hint, set 1/u=x/n, and re-express in terms of u and x.

5. Mar 15, 2006

### island-boy

which one of the limits are wrong?

I was able to prove that:
$$(1 + x/n)^n \leq e^x$$ and $$(1 - x/n)^n \leq e^{-x}$$

Using the Taylor expansion for $$e^x$$, I got

$$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + ...$$
which means
$$e^x \geq 1 + x$$
subtituting x by x/n, I get
$$e^{x/n} \geq 1 + x/n$$
thus
$$e^{(x/n)n} \geq (1 + x/n)^n$$
simplifying, I get:
$$e^x \geq (1 + x/n)^n$$

Similarly I used the Taylor Expnansion to prove that
$$e^x = \lim_{n\rightarrow\infty} (1+x/n)^n$$

since $$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} +...$$

and using binomial theorem for $$\lim_{n\rightarrow\infty} (1+x/n)^n$$
I got:
$$\lim_{n\rightarrow\infty}1^n + \lim_{n\rightarrow\infty} x + \lim_{n\rightarrow\infty} \frac{n(n-1)(x^2)}{(n^2)(2!)} + \lim_{n\rightarrow\infty} \frac{n(n-1)(n-2)(x^3)}{(n^3)(3!)} + \lim_{n\rightarrow\infty} \frac{n(n-1)(n-2)(n-3)(x^4)}{(n^4)(4!)} + ... = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} +... =e^x$$

these correct?

Last edited: Mar 15, 2006
6. Mar 15, 2006

### Hurkyl

Staff Emeritus
You've used n-->infinity in expressions that don't even have an n in them! (You meant h-->0)

7. Mar 15, 2006

### benorin

See this thread posts # 3 and 12.

I take it that your definition for $$e^x$$ is the Taylor series about x=0, viz.

$$e^x=\sum_{k=0}^{\infty}\frac{x^k}{k!}$$

Last edited: Mar 15, 2006
8. Mar 15, 2006

### benorin

Essentially, this

and this,