# Limit prove simple

1. Nov 20, 2009

### andilus

how to prove
lim(x->a)$$\frac{f(x)-f(a)}{x-a}$$=lim(h->0)$$\frac{f(a+h)-f(a)}{h}$$

it seems to be obvious, but i dunno how to prove`

2. Nov 20, 2009

What is the relationship between the denominator $$x - a$$ and the denominator $$h$$?

3. Nov 20, 2009

### andilus

Sorry,I have not express clearly.
what i want to prove is just:
lim(x->a)$$(\frac{f(x)-f(a)}{x-a})$$=lim(h->0)$$(\frac{f(a+h)-f(a)}{h})$$

4. Nov 20, 2009

### l'Hôpital

Listen to what statdad is saying. What is the relationship by h and x - a? The answer to this question will essentially answer your question.

5. Nov 20, 2009

### wisvuze

use a delta-epsilon argument, the standard delta epsilon definition:

as x approaches a, we have:

given $$\epsilon$$ > 0, there exists a $$\delta$$ >0 such that for all x with the property 0 < | x - a | < $$\delta$$, then |f(x) - L | < $$\epsilon$$.

In this argument, we have the distance between a point x and a fixed point a bounded between 0 and some fixed $$\delta$$. Can you provide a similar argument as h approaches ____ ?

6. Nov 20, 2009

### Bohrok

Why not let h = x - a and rewrite the limit after the substitutions?

7. Nov 21, 2009

### HallsofIvy

Staff Emeritus
That is exactly what statdad was suggesting!

8. Nov 21, 2009

### Bohrok

I wasn't sure that that was what statdad was getting at, which is why I posted the equation. Some things are just too subtle, at least for me.

9. Nov 22, 2009

### HallsofIvy

Staff Emeritus
You still don't get it? In the first limit,
$$\lim_{x\to a}\frac{f(x)- f(a)}{x-a}$$

let h= x- a. Then x= ??

10. Nov 22, 2009

### Staff: Mentor

I'm pretty sure none of this is applicable to the problem in this thread.

11. Nov 22, 2009

### Bohrok

No, I understand perfectly how to use h = x - a to transform the first to the second; I just didn't see that statdad was hinting at doing it that way.

12. Nov 22, 2009