# Proof of a limit

1. Oct 29, 2007

### yhsbboy08

can somebody explain to me how to prove this:

the limit, as h approaches 0, of ((e^h)-1)/h equals 1.

sorry about the typing, idk how else to get it on here...

2. Oct 29, 2007

### PowerIso

Do you want a real proof? Or just how you get the answer?

use L'h rule you get e^h/1 take the limit and you get 1

but as for a formal proof, i'm to tired to think lol.

3. Oct 29, 2007

### d_leet

L'hopitals rule is not a way to do this because to do that one usually has to know the derivative of e^x, but this limit is, itself, needed in order to figure out this derivative. So for a better proof try using the definition of e, this is not particularly hard if you use the definition of e as the limit (1+h)^(1/h) as h approaches zero.

4. Oct 29, 2007

### yhsbboy08

can u explain it a lil further? what do you mean by the definition of e?

5. Oct 29, 2007

### d_leet

Can you explain it to me? What do you mean when you use e? It is obviously a number, but usually it is defined in some way when one is first introduced to it, in your course or textbook or whatever how is it defined?

6. Oct 29, 2007

### robert Ihnot

I think he wants you to look at it this way: $$e^h=\sum_{j=0}^\infty\frac{h^j}{j!}=1+h+\frac{h^2}{2!} ++++$$ Now plug that into your original equation.

Last edited: Oct 29, 2007
7. Oct 29, 2007

### d_leet

Actually I was aiming for e= the limit as n approaches infinity of [1+(1/n))]^n which is equivalent to the limit as h approaches 0 of (1+h)^(1/h)

8. Oct 30, 2007

### robert Ihnot

Well, regardless, if you can make that work out to something like: 1+1+(1-h)/2! +(1-h)(1-2h)/3!+++ all you arrive at as h goes to zero is the value for e, not the value for e^h.

Now when I took the calculus, I believe e^x was defined as $$lim_{n\rightarrow\infty} (1+x/n)^n$$

9. Oct 30, 2007

### d_leet

All you need is the value of e though.

10. Oct 30, 2007

### robert Ihnot

You mean $$\frac{[(1+h)^{1/h}]^h-1}{h}$$?

Last edited: Oct 30, 2007
11. Oct 30, 2007

Exactly.