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Is this proof (about limits) acceptable?

  1. Mar 24, 2012 #1
    This is not homework.
    Earlier today I was trying to prove that if a limit of a certain function exists, then it's unique:

    limf(x)=a [itex]\wedge[/itex] limf(x)=b (as x→x0) then a=b

    I began to use the sum of limits like so:

    limf(x)+limf(x)=a+a → lim2f(x)=2a (as x→x0)

    And the same thing for limf(x)=b results in lim2f(x)=2b.

    Now, I thought that if limf(x)=a [itex]\wedge[/itex] limf(x)=b, then:

    lim2f(x)=a+b (as x→x0)

    I concluded that a+b=2b [itex]\vee[/itex] a+b=2a, which gave me a=b on both equations.

    Is this proof acceptable or do I have to prove it by the definition of limit?
  2. jcsd
  3. Mar 24, 2012 #2


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    Not acceptable: how did you conclude [itex]a+b = 2a \vee a+b = 2b[/itex]? It looks like you used the very thing you were trying to prove!
  4. Mar 24, 2012 #3
    Hum, can you explain me how is this redundant? I'm not seeing it.. '.'
    I simply summed both limits.
  5. Mar 24, 2012 #4


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    Dearly Missed

    Well, you assume, for example, you can sum limits, and also split them up as you choose among the addends.

    Why do you believe this would be the case if the limit was a NON-unique number?

    Instead, think about the following a bit, with "a" and "b" as possibly different limits.

    |a-b|=|a-f(x)+f(x)-b|<=|a-f(x)|+|b-f(x)|, by the triangle inequality, valid for any and every value of "x" you might choose to consider.

    Now, how can you proceed from here to conclude that the absolute value of the difference between the two fixed numbers "a" and "b" must be less than any chosen positive number you can think of?
    Last edited: Mar 24, 2012
  6. Mar 24, 2012 #5


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    You proved

    lim 2f(x) = 2a
    lim 2f(x) = 2b
    lim 2f(x) = a+b

    and now you're trying to make statements like 2a = a+b.

    When you're proving things like this, it might help to write things as a relation that you would normally write with a fake equality symbol, so you don't confuse yourself by expecting transitivity. i.e. write

    L(f(x), a)​

    for what you would normally write as

    lim f(x) = a.​

    So you're trying to prove

    L(f(x), a) and L(f(x), b) together imply a = b​

    (incidentally, this is the theorem that justifies writing lim f(x)=a. Technically you should never write that notation until you know this theorem)

    and your working has shown that

    L(2f(x), 2a) and L(2f(x), a+b) and L(2f(x),2b)​

    e.g. by invoking the theorem

    L(g(x), u) and L(h(x), v) together imply L(g(x) + h(x), u+v)​

    Now, how do you plan to conclude anything from those three relations?
    Last edited: Mar 24, 2012
  7. Mar 24, 2012 #6


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    Dearly Missed

    To expand upon Hurkyl's comment:
    What do you really KNOW about the number "a" that is claimed is the limit of f as x is close enough to x0?

    What you KNOW is that the absolute value of the difference, |a-f(x)| can be made as small as you wish, as long as you let x be close enough to x0.

    THAT is what you know, and nothing else. So that nugget of knowledge is all you are allowed to play around with in constructing your proof.
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