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Math Help - All the cases in which a limit cannot exist

  1. #1
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    All the cases in which a limit cannot exist

    My prof said there was three ways, so far I've found two:

    When the limit goes up to infinity, and when the left and right limits don't match. What are the other ways a limit cannot exist?
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  2. #2
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    Quote Originally Posted by AMaccy View Post
    My prof said there was three ways, so far I've found two:

    When the limit goes up to infinity, and when the left and right limits don't match. What are the other ways a limit cannot exist?
    An example for you to consider: \lim_{x \rightarrow 0} \sin \left(\frac{1}{x}\right).
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  3. #3
    Senior Member pankaj's Avatar
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    There are indeed three ways in which the \lim_{x\to a}f(x) cannot exist.

    1. f(x) is not well defined inthe neighbourhood of x=a.

    For example \lim_{x\to 1}\sec^{-1}(\sin x) does not exist

    since \sec^{-1}(\sin x) is not defined in the neighbourhood

    of x=1 though it is defined at x=1.The same

    is true for \lim_{x\to 1}\sin^{-1}(\frac{1+x^2}{2x}).

    2. f(x) does not display tendency to approach a fixed numerical value.

    For e.g. it is not known as to what value

    \lim_{x\to 0}\sin (\frac{1}{x}) will assume though we are

    sure that it is a finite quantity on the interval [-1,1].

    Similarly, f(x)=\begin{cases}0,\mbox { if } x\in irrationals\\1,\mbox{ if } x\in rationals\end{cases}

    Here, f(3) = 1.But if you try to evaluate \lim_{x\to 3}f(x) we will not be able to decide to what value should f(x)

    assume on the interval [3-h,3+h]\notin {3}.

    3.Of course ,the last condition being Left hand limit \neq Right hand limit
    Last edited by pankaj; June 8th 2009 at 03:20 AM.
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