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Epsilon delta proofs equaling a constant

  • #1

Homework Statement



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

Homework Equations





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!
 

Answers and Replies

  • #2
LCKurtz
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Homework Statement



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

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!
I have no idea either. What is your question? What are you trying to do?
 
  • #3
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
SammyS
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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.
δ can be pretty much anything.

Let δ = 1, for example.
 
  • #5
I understand that but my proff wants us to mathematically prove why that is. Wich seems difficult because it is s simple.
 
  • #6
HallsofIvy
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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)."
 
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  • #7
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?
 
  • #8
No, what I wrote can't be right.. Uh
 
  • #9
SammyS
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No, what I wrote can't be right.. Uh
How about:

For ε > 0

If | x - a | < δ

then | f(x) - c | =   ?  
 
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  • #10
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
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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.
Yes. Don't overthink it.
Can someone just post the solution so I can follow the steps and really learn this?
Not going to happen.

See the rules (https://www.physicsforums.com/showthread.php?t=414380), especially this one:
On helping with questions: Any and all assistance given to homework assignments or textbook style exercises should be given only after the questioner has shown some effort in solving the problem. If no attempt is made then the questioner should be asked to provide one before any assistance is given. Under no circumstances should complete solutions be provided to a questioner, whether or not an attempt has been made.
 
  • #12
Guess I should have read that first. It has to be l fx-c l < δ that's what I'm thinking
 
  • #13
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Guess I should have read that first. It has to be l fx-c l < δ that's what I'm thinking
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
SInce l fx-L l = 0 and ε > 0 then l fx - L l < ε. Please be right....
 
  • #15
LCKurtz
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SInce l fx-L l = 0 and ε > 0 then l fx - L l < ε. Please be right....
And notice that, for this particular function, it doesn't matter how big or small ##\delta## is.
 
  • #16
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
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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
You really should get into the habit of writing function notation correctly. It's f(x), not fx.
 
  • #18
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Here an example
http://imageshack.us/a/img708/1055/10204791.jpg [Broken]
 
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