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

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

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

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- #4

SammyS

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δ can be pretty much anything.

Let δ = 1, for example.

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- #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)."

"Given [itex]\epsilon> 0[/itex], let [itex]\delta[/itex] be

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

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SammyS

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

For ε > 0

If | x - a | < δ

then | f(x) - c | =

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Mark44

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Yes. Don't overthink it.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.

Not going to happen.Can someone just post the solution so I can follow the steps and really learn this?

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.

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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

- #13

Mark44

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

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

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And notice that, for this particular function, it doesn't matter how big or small ##\delta## is.SInce l fx-L l = 0 and ε > 0 then l fx - L l < ε. Please be right....

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- #17

Mark44

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

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Here an example

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

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

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