Epsilon delta proofs equaling a constant

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!

LCKurtz

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

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.

SammyS

δ can be pretty much anything.

Let δ = 1, for example.

tachyon_man

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

HallsofIvy

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

tachyon_man

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

tachyon_man

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

SammyS

How about:

For ε > 0

If | x - a | < δ

then | f(x) - c | =   ?  

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?

Mark44

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

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

tachyon_man

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

Mark44

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.

tachyon_man

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

LCKurtz

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

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

Mark44

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