Epsilon delta proofs equaling a constant

In summary, the conversation revolves around proving that as x approaches a, the limit of a constant function is equal to the constant. The participants discuss various methods and equations, and eventually arrive at the conclusion that since the function f(x) is constant, its limit as x approaches a will also be constant. The conversation also emphasizes the importance of showing effort and not providing complete solutions to homework problems.
  • #1
tachyon_man
50
0

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!
 
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  • #2
kylem1994 said:

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
kylem1994 said:
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
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
kylem1994 said:
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
kylem1994 said:
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.
kylem1994 said:
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
kylem1994 said:
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
kylem1994 said:
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
kylem1994 said:
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
Here an example
http://imageshack.us/a/img708/1055/10204791.jpg
 
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FAQ: Epsilon delta proofs equaling a constant

1. What is an epsilon delta proof?

An epsilon delta proof is a mathematical method used to prove the limit of a function. It involves choosing two values, epsilon (ε) and delta (δ), and showing that for any input that is within δ distance of the limit, the output will be within ε distance from the limit. This is used to rigorously prove that a limit exists and has a specific value.

2. How is an epsilon delta proof used to show that a function equals a constant?

In an epsilon delta proof, we start by assuming that the limit of the function is equal to the constant we are trying to prove. Then, we use the definition of a limit to choose appropriate values for ε and δ, and show that for any input within δ distance from the limit, the output will be within ε distance of the constant. This proves that the function equals the constant at that specific point.

3. Why is it important to use an epsilon delta proof to show that a function equals a constant?

Using an epsilon delta proof adds rigor and precision to mathematical proofs. It allows us to prove the existence and value of a limit in a formal and conclusive way. Additionally, it helps to eliminate any potential errors or assumptions that may occur in less rigorous methods of proof.

4. What are the limitations of using an epsilon delta proof to show that a function equals a constant?

An epsilon delta proof can be quite complex and time-consuming, requiring a deep understanding of the definition of limits and mathematical concepts. It may also be difficult to apply this method to more complex functions. Additionally, it only proves that a function equals a constant at a specific point, and not for the entire domain of the function.

5. How can one improve their skills in using an epsilon delta proof to show that a function equals a constant?

Practice is key in improving skills in using epsilon delta proofs. It is also important to have a strong understanding of mathematical concepts such as limits and continuity. Working through examples and seeking help from professors or tutors can also aid in improving skills in this area.

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