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Proving the limit does not exist formally

  1. Oct 17, 2013 #1
    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. jcsd
  3. Oct 17, 2013 #2


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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.
  4. Oct 17, 2013 #3
    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
    $$\exists \epsilon > 0\ \forall\ \delta > 0\ , there\ is\ x\ satisfying\ 0< |x-c| < \delta \ but\ |f(x) - l| >= \epsilon$$
  5. Oct 17, 2013 #4
    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.
  6. Oct 17, 2013 #5
    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
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