Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Proof of uniqueness of limits

  1. Feb 7, 2010 #1
    1. The problem statement, all variables and given/known data Can anyone help me with proving the uniqueness of a limit? The one that stated that a limit, L, only exists if the left and right hand limits at that point are the same?

    2. Relevant equations

    3. The attempt at a solution
    I started by saying that let us say a function f(x) has two limits, L1 and L2 at the point a, such that L1<L2, and there exists for both the same epsilon and delta.

    As x →a-, lim x→a of f(x)= L1,

    Such that |f(x)-L1|<epsilon , which implies 0<|x-a|<delta ………..(1)

    Then as x →a+, lim x→a of f(x)=L2

    Such that |f(x)-L2|<epsilon, which implies 0<|x-a|<delta ……….(2)

    By subtracting the epsilon statements from each other, I am left with:
    0<L1-L2<0, which is a contradiction, hence L1 and L2 must be the same.

    I don't know if this is a correct method of proving this, so I would greatly appreciate feedback. If there are any other methods, I would greatly appreciate it. Sorry I couldn't use all the proper mathematical symbols.
    Last edited: Feb 7, 2010
  2. jcsd
  3. Feb 7, 2010 #2
    This is a fairly basic question so you're probably in introductory real analysis or a calculus course. At places you are omitting parts of the argument, and I don't know whether this is due to a lack of understanding, but I would urge you to make your arguments very explicit when you're getting familiar with proofs.

    The first thing you do is assume that the limit of f(x) as x goes to a approaches L1 and L2. This is correct, but then you make some vague statement about epsilons and deltas and you state the converse statements of convergence which is an error. What you really should do is be explicit and careful about it. For every [itex]\epsilon>0[/itex] since f(x)->L1 there exists some [itex]\delta_1 > 0[/itex] such that [itex]0 < |x-a| < \delta_1[/itex] imply [itex]|f(x)-L_1| < \epsilon[/itex]. Similarily since f(x)->L2 there exists some [itex]\delta_2 > 0[/itex] such that [itex]0 < |x-a| < \delta_2[/itex] imply [itex]|f(x)-L_2| < \epsilon[/itex]. As you rightly observed we would really like only one delta since we want to consider both inequalities. Thus what we do is set [itex]\delta = \min(\delta_1,\delta_2)[/itex]. Then [itex]\delta>0[/itex] and [itex]0 < |x-a| < \delta[/itex] imply [itex]0 < |x-a| < \delta \leq \delta_1[/itex] and similarily [itex]0 < |x-a| < \delta \leq \delta_2[/itex].

    To summarize you now know that given any [itex]\epsilon>0[/itex], there exists some [itex]\delta>0[/itex] such that if [itex]0 < |x-a| < \delta[/itex], then [itex]|f(x)-L_1| < \epsilon[/itex] and [itex]|f(x) -L_2| < \epsilon[/itex]. You did some weird subtraction of absolute values that isn't valid (you can't just subtract inequalities from eachother and |x-y| = |x|-|y| is not true in general). What you really need to do is note:
    [tex]2\epsilon > |f(x)-L_2| + |L_1-f(x)|[/tex]
    Then apply the triangle inequality to get:
    [tex]2\epsilon > |f(x)-L_2| + |L_1-f(x)| \geq |f(x)-L_2 + L_1 - f(x)| = |L_1-L_2|[/tex]
    [itex]\epsilon[/itex] was an arbitrary positive value so if L1 differ from L2, choose it to be:
    [tex]\epsilon= |L_1-L_2|/2[/tex]
    Then you get,
    [tex]2\epsilon = |L_1-L_2| > |L_1-L_2|[/tex]
  4. Feb 7, 2010 #3
    Thank you for the detailed explanation. Indeed, it seems that at the end you obtained a contradiction as well? Just asking for clarification. Thank you again.
  5. Feb 7, 2010 #4
    Yes. x>x is always a contradiction so our assumption that L1 differed from L2 must have been false.
  6. Feb 7, 2010 #5
    Thanks again, and one last question. The proof you detailed indeed covers the condition that left and right hand limits must be the same if the limit exists at a point right? I still am not sure if uniqueness of a limit is the same as saying that the left and right hand limits must be the same.
  7. Feb 7, 2010 #6
    It doesn't cover it explicitly, but it could fairly easily be adapted to prove it as the idea is identical. Simply assume L1 is the two-sided limit and that L2 is a limit from the left side and show L1=L2 in the same way (you will need to replace [itex]0 < |x-a| < \delta_2[/itex] by [itex]0 < a-x < \delta_2[/itex], but this implies the former so we can follow the proof in the previous post). You can do the right hand side in exactly the same way.
  8. Feb 7, 2010 #7
    I'll try that, thanks
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook