Find the Condition for Limit of f(x) Not Equal to L | Given \epsilon > 0

[jdstokes]

Let f: \mathbb{R} \rightarrow \mathbb{R} be given. Let L be a real number. State the condition for saying that as x tends to a, the limit of f(x) is not L. The statement ought to begin with "Given there exists \epsilon > 0".

Best guess: \lim_{x \rightarrow a}f(x) \neq L means, given there exists \epsilon > 0 there exists \delta > 0 such that |x-a|<\delta \Rightarrow |f(x) - L| > \epsilon. I'm really not sure about this, however.

 

[whozum]

Nonremovable singularity?

 

[jdstokes]

Pardon me?

 

[saltydog]

First, let's go over the rule for the limit:

Given \epsilon>0, there exists a \delta>0 such that for all x satisfying:

|x-a|<\delta

we have:

|f-L|<\epsilon


Now, what happens if L is not the limit?


Then there should exists an \epsilon>0 such that for some \delta>0 and all x satisfying:


|x-a|<\delta

we have:

|f-L|>\epsilon

Make sure you understand these two differences. If L is the limit, by choosing \delta small enough I can make the difference between L and f as small as I want.

If L is not the limit, then no matter how small I make \delta, I can always find an \epsilon such that the difference between f and L will be larger.

 

[jdstokes]

So I was correct? Great, thanks for your help saltydog.