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Homework Help: Why we use strictly less than delta and epsilon in definition of limits

  1. Oct 5, 2012 #1
    1. The problem statement, all variables and given/known data

    I'm wondering why we can't use less than or equal to for the formal definition of the limit of a function:

    2. Relevant equations

    lim x→y f(x)=L iff For all ε>0 exists δ>0 such that abs(x-y)<δ implies abs(f(x) - L)<ε

    Why not:

    lim x→y f(x)=L iff For all ε>0 exists δ>0 such that abs(x-y)=<δ implies abs(f(x) - L)=<ε

    3. The attempt at a solution

    I just imagine that it doesn't matter because epsilon is arbitrary, but I'm not sure. If we're using the epsilon such that we have a strict inequality, given that set of x that satisfies this, all we're doing is including the boundary points of the open set specified by delta on the domain, and we know that by the definition using strict inequalities that there then exists a [ε][/1]>ε such that abs(f(x) - L)<[ε][/1] (and thus abs(f(x) - L)=<[ε][/1], but we know that there are not x in our ball that actually satisfy abs(f(x) - L)=[ε][/1]).

    But I can't seem to find a way to satisfy a two way implication i.e. prove the definitions are equivalent. Maybe there's a good contradiction up someone knows two show they're not equivalent?
  2. jcsd
  3. Oct 5, 2012 #2


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    The definitions are equivalent.

    Suppose f satisfies the <= definition of [itex]\lim_{x \rightarrow y}f(x) = L[/itex]. Let [itex]\epsilon > 0[/itex]. Then there exists [itex]\delta > 0[/itex] such that [itex]0 < |x - y| \leq \delta[/itex] implies [itex]|f(x) - L| \leq \epsilon/2[/itex]. Thus clearly [itex]0 < |x - y| < \delta[/itex] also implies [itex]|f(x) - L| \leq \epsilon/2[/itex]. But [itex]\epsilon/2 < \epsilon[/itex], so f satisfies the < definition of [itex]\lim_{x \rightarrow y}f(x) = L[/itex].

    You can prove the converse quite similarly.
  4. Oct 6, 2012 #3
    Thanks - I was having trouble trying to find a way to imagine picking a larger epsilon while keeping it arbitrarily close to zero, but I see now that if we define the relation as a proportion of the smaller epsilon, this works. Thanks.
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