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Uniqueness of Limits Proof

  1. Jan 20, 2013 #1
    1. The problem statement, all variables and given/known data
    Using the delta-epsilon definition of limits, prove that of lim f(x) = l and lim f(x) =m, then l=m


    2. Relevant equations
    Delta-epsilon definiition of the limit of f(x), as x approaches a:
    For all e>0, there is a d s.t if for all x, |x-a|<d, then |f(x) -l|<e

    3. The attempt at a solution
    Suppose not. Then
    (1): For all e>0, there is a d1 s.t. |f(x) - l|<e, provided |x-a|<d1 for all x
    (2): For all e>0, there is a d2 s.t. |f(x) - m|<e, provided |x-a|<d2 for all x

    I'm getting stuck at finding the delta that works for both of them. I see that the next step is to pick d=min(d1, d2), but I'm confused as to why this is the case

    For instance, if d1=1, and d2=2, I don't see how d=1 would satisfy (2) above
     
  2. jcsd
  3. Jan 20, 2013 #2
    Given some ##\delta## that works in an epsilon-delta proof, you can always take a smaller delta, ##\delta'##, instead (since ##|x-a|<\delta' < \delta \rightarrow |x-a| < \delta##). So if ##\delta_1## works for (1) and ##\delta_2## works for (2), then ##min(\delta_1,\delta_2)## works for both.
     
  4. Jan 20, 2013 #3
    Alright, I see. I was thinking max and confusing myself, lol. Thanks!
     
    Last edited: Jan 20, 2013
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