# Proving the limit does not exist formally

1. Oct 17, 2013

### NATURE.M

So I've searched around quite a bit, and have been fairly unsuccessful when it comes to finding any sufficient material on disproving a limit using the epsilon delta def. I was wondering if any of you could recommend any good sources for learning how to disprove a limit using epsilon delta.

2. Oct 17, 2013

### arildno

Both proving and disproving specific limits, for example to come up with narrow error estimates of the limit value, is generally an extremely tricky business.

3. Oct 17, 2013

### DarthRoni

Recall the definition of $\lim_{x\rightarrow c}f(x) = l$
$$\forall \epsilon > 0\ \exists\ \delta > 0 \ such\ that\ 0<|x-c|<\delta \implies |f(x) - l| < \epsilon$$
Say that is it not true that $lim_{x\rightarrow c}f(x) = l$ this means
$$\exists \epsilon > 0\ \forall\ \delta > 0\ , there\ is\ x\ satisfying\ 0< |x-c| < \delta \ but\ |f(x) - l| >= \epsilon$$

4. Oct 17, 2013

### NATURE.M

I understand the negation of the epsilon-delta statement, but you could say is there a particular way to approach such a task. When your trying to find an epsilon is there a general procedure you could perform. Or does it really just always change from limit to limit.

5. Oct 17, 2013

### DarthRoni

Here's an example:
Disprove $\lim_{x \rightarrow c}x+2 = c$
we want to show for some $\epsilon > 0$ then $\forall \delta>0$
$$|x-c|<\delta \implies |(x+2) - c| >= \epsilon$$
We know
$$|(x+2) - c| = |(x-c) + 2|$$
And let's say $$|x-c|<1$$
Because remember that we are trying to show that there is an x in which the definition falls apart. So,
$$-1 < x -c < 1 \implies |(x-c) + 2| > |-1 + 2| = 1$$
So we can say that any $\epsilon < 1$ works. Notice that I chose |x-c| to be smaller than 2.
HINT: really any $\epsilon < 2$ works
EDIT: Just as arildno said, theses proofs can get REALLY tricky.

Last edited: Oct 17, 2013