# Epsilon delta proofs equaling a constant

1. Sep 14, 2012

### tachyon_man

1. The problem statement, all variables and given/known data

Lim x→a of f(x) = c (Where c is a constant)

2. Relevant equations

3. The attempt at a solution
I have no idea. I am able to do these if I can manipulate fx-L to equal x-a but I am having trouble with this one. Please help me!

2. Sep 14, 2012

### LCKurtz

I have no idea either. What is your question? What are you trying to do?

3. Sep 14, 2012

### tachyon_man

I have to prove using the epsilon delta method that as x approaches a the limit of a constant is a constant. It seems so easy to do that I'm having trouble with it.

4. Sep 14, 2012

### SammyS

Staff Emeritus
δ can be pretty much anything.

Let δ = 1, for example.

5. Sep 14, 2012

### tachyon_man

I understand that but my proff wants us to mathematically prove why that is. Wich seems difficult because it is s simple.

6. Sep 14, 2012

### HallsofIvy

Staff Emeritus
You are claiming that for f(x)= c, a contant, for any a, $\lim_{x\to a} f(x)= c$. The definition of "$\lim_{x\to a} f(x)= L$" is "Given $\epsilon> 0$, there exist $\delta> 0$ such that if $|x- a|< \delta$ then $|f(x)- L|< \epsilon$". Here, of course, f(x)= c and L= c so that $|f(x)- L|= |c- c|= 0<\epsilon$ for all x. So say
"Given $\epsilon> 0$, let $\delta$ be any positive number. Then if $|x- a|< \delta$... (you do the rest)."

Last edited by a moderator: Sep 14, 2012
7. Sep 14, 2012

### tachyon_man

Thank you very much :) after which I say since $\left|x-a \right|$ = $\left| fx-L \right|$ then $\left| fx-L \right|$ > δ. And that completes the proof?

8. Sep 14, 2012

### tachyon_man

No, what I wrote can't be right.. Uh

9. Sep 14, 2012

### SammyS

Staff Emeritus

For ε > 0

If | x - a | < δ

then | f(x) - c | =   ?

Last edited: Sep 14, 2012
10. Sep 14, 2012

### tachyon_man

All the comments have been very helpful and I'm very embarrassed because I'm very good at math.. I think I'm really overthinking it. Can someone just post the solution so I can follow the steps and really learn this?

11. Sep 14, 2012

### Staff: Mentor

Yes. Don't overthink it.
Not going to happen.

See the rules (https://www.physicsforums.com/showthread.php?t=414380), especially this one:

12. Sep 14, 2012

### tachyon_man

Guess I should have read that first. It has to be l fx-c l < δ that's what I'm thinking

13. Sep 14, 2012

### Staff: Mentor

No, you want to show that |f(x) - C| < $\epsilon$. That really shouldn't be hard to do, given the function you're working with.

14. Sep 14, 2012

### tachyon_man

SInce l fx-L l = 0 and ε > 0 then l fx - L l < ε. Please be right....

15. Sep 14, 2012

### LCKurtz

And notice that, for this particular function, it doesn't matter how big or small $\delta$ is.

16. Sep 14, 2012

### tachyon_man

Yeah it finally clicked. If l fx - L l is 0 and δ is any positive number, Given the information, it fx-L must be less than

17. Sep 14, 2012

### Staff: Mentor

You really should get into the habit of writing function notation correctly. It's f(x), not fx.

18. Sep 14, 2012

### azizlwl

Here an example
http://imageshack.us/a/img708/1055/10204791.jpg [Broken]

Last edited by a moderator: May 6, 2017