# Proving the limit does not exist formally

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.

Homework Helper
Gold Member
Dearly Missed
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.

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
$$\exists \epsilon > 0\ \forall\ \delta > 0\ , there\ is\ x\ satisfying\ 0< |x-c| < \delta \ but\ |f(x) - l| >= \epsilon$$

NATURE.M
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
$$\exists \epsilon > 0\ \forall\ \delta > 0\ , there\ is\ x\ satisfying\ 0< |x-c| < \delta \ but\ |f(x) - l| >= \epsilon$$

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.

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.

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