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- Thread starter NATURE.M
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arildno

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$$\forall \epsilon > 0\ \exists\ \delta > 0 \ such\ that\ 0<|x-c|<\delta \implies |f(x) - l| < \epsilon$$

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

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$$\forall \epsilon > 0\ \exists\ \delta > 0 \ such\ that\ 0<|x-c|<\delta \implies |f(x) - l| < \epsilon$$

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.

- #5

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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.

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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