Proving the limit does not exist formally

  • Thread starter NATURE.M
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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.
 

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  • #2
arildno
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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
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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
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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.
 
  • #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.
 
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