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Homework Help: I am realizing these proofs are meaningless!

  1. Jan 3, 2012 #1
    I've been looking at this proof thinking that if I read it over and over that what I am reading that seems so obvious that something else will actually pop out that I am not realizing, but what I realized the proofs that I am reading seem meaningless and pointless.

    I added a paint doc with the theorem, then the proof.. then my thoughts at the end... I want to know if my thoughts are correct?

    thank you.

    Attached Files:

    • PQ.jpg
      File size:
      43.3 KB
  2. jcsd
  3. Jan 3, 2012 #2

    Ray Vickson

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    I think the proof is off base, because it is looking at the deltas instead of the epsilons. Basically, you need to ensure that ε is small enough that (for l ≠ m) the intervals (l-ε, l+ε) and (m-ε, m+ε) are non-overlapping; then it would be impossible for all f(x) (x near a) to be in both intervals at the same time.

  4. Jan 3, 2012 #3
    Thats another thing... if they overlapped then that would go against the definition of a limit. I also thought of that but I was thinking that is to obvious by the def of limit
  5. Jan 3, 2012 #4
    Is that all proofs are is stating the obvbious?
  6. Jan 3, 2012 #5


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    Your attachment seems to be an incomplete proof - you haven't got to the "punchline" yet.

    You will find many things in analysis (and the rest of math!) that look obvious but are actually false. There are also some very "non-obvious" things that are actually true.

    That's why theorems like this need to be PROVED, starting from the definition of a limit, not just "assumed to be true because they look obvious."
  7. Jan 3, 2012 #6
    Sorry I must have cut the last part out... here is the last bit of it.

    But this last part seems pointless to me also, but obviously its not if it was pointed out that something was missing...

    I say it seems pointless because its obvious f(x) can not approach 2 diff limiting values neer one and only one x value at a.

    So basically to me this last part has the SAME exact meaning as the above paragraph just stated differently.. the above paragraph seems to be proving directly that there can not be two limits by picking the min delta value, then from there restating the limit definition... but the second part that I posted seems to prove it indirectly by contradiction.

    And I am saying they look obvious just by the definition of limit. I am sure if I didnt know the definition and was trying to understand the proof first it wouldn't seem so obvious...

    Attached Files:

    • R.jpg
      File size:
      17.6 KB
  8. Jan 3, 2012 #7


    Staff: Mentor

    The part you left off is the contradiction. The assumption was that lim f(x) = l and lim f(x) = m, and that l ≠ m (which means that |l - m| > 0).

    Starting with the assumption that lim f(x) = l and lim f(x) = m, they arrived at |l - m| < |l - m|, which cannot be true. A number cannot be less than itself.
  9. Jan 3, 2012 #8
    I know thats what I said, the part I left out was the contradiction... but i feel like the second part I posted is saying the same exact thing as the first thing I posted in a different way.
  10. Jan 3, 2012 #9
    No its not..
    So this proof may seem fairly obvious but you have to prove all the basic stuff once so that later on you can just write 'by uniqueness of limits' whenever you please.

    The proof is basically assuming that somehow a function tends to 2 different things and then reaches a contradiction. The first half just says that when x gets sufficiently close to a, f(x) gets close to l and f(x) also gets close to m. The contradiction bit then shows it is impossible for l and m to then be different (which is what Ray was saying in the first post). So it is definitely needed for the proof :)
  11. Jan 4, 2012 #10
    Thanks, makes sense. I was actually thinking something else that was incorrect on my part but you cleared it up.
  12. Jan 4, 2012 #11


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    well, no, they're not quite meaningless. consider the following function:

    f(x) = 0, x rational.
    f(x) = 1, x irrational.

    what is the behavior of this function "near 1"?

    the important part of the theorem, and one that is often overlooked (because of its simplicity), is that f HAS a limit at a. this kind of behavior is special, not all functions behave so nicely.
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