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.