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Math Help - limit proof

  1. #1
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    limit proof

    Hi all,

    Just wondering if I could get help with the following problem.

    The question states that I have to prove that:

    if lim(x->c) f(x) = L, then lim(x->c)|f(x)| = |L|

    I've already proven the above. But then the question goes onto saying.

    Show that the converse if false. Give an example where:

    lim(x->c)|f(x)| = |L| and lim(x->c)f(x) = M where M is not equal to L.

    Also, it asks to give an example where

    lim(x->c)|f(x)| exists but lim(x->c)f(x) does not exist.

    I'm a bit stumped with this, I can't think of one simple example. Any help would be appreciated.

    Cheers.
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  2. #2
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    Quote Originally Posted by mosostow View Post
    Hi all,

    Just wondering if I could get help with the following problem.

    The question states that I have to prove that:

    if lim(x->c) f(x) = L, then lim(x->c)|f(x)| = |L|

    I've already proven the above. But then the question goes onto saying.

    Show that the converse if false. Give an example where:

    lim(x->c)|f(x)| = |L| and lim(x->c)f(x) = M where M is not equal to L.

    Also, it asks to give an example where

    lim(x->c)|f(x)| exists but lim(x->c)f(x) does not exist.

    I'm a bit stumped with this, I can't think of one simple example. Any help would be appreciated.

    .
    How about sgn(x)
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  3. #3
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    Here is another one I like.
    Suppose that f(x)=1 if x is rational and f(x)=-1 if x is irrational. Now |f(x)| has a limit everywhere and it is 1; but f(x) has a limit nowhere.
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  4. #4
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    Quote Originally Posted by Plato View Post
    Here is another one I like.
    Suppose that f(x)=1 if x is rational and f(x)=-1 if x is irrational. Now |f(x)| has a limit everywhere and it is 1; but f(x) has a limit nowhere.
    I just love the Dirichlet and his function. I like him so much that I managed to somehow mention his name in my English composition essay.
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  5. #5
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    Quote Originally Posted by Plato View Post
    Here is another one I like.
    Suppose that f(x)=1 if x is rational and f(x)=-1 if x is irrational. Now |f(x)| has a limit everywhere and it is 1; but f(x) has a limit nowhere.
    Thanks, this answers the second part of the question.

    But I'm still a bit stumped with the first part, that is, to give one example where

    if lim(x->c)|f(x)| = |L| then lim(x->c)f(x) = M where M not equal to L.

    To me it seems like a logical absurdity but apparently an example does exist.
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  6. #6
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    What about f(x) = -1 then lim (x-->c)f(x)=-1

    But, lim (x-->c)|f(x)| = 1

    And 1 != -1
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