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Thread: 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: $\displaystyle \lim_{x \rightarrow 0} \sin \left(\frac{1}{x}\right)$.
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  3. #3
    Senior Member pankaj's Avatar
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    There are indeed three ways in which the $\displaystyle \lim_{x\to a}f(x)$ cannot exist.

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

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

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

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

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

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

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

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

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

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

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

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

    3.Of course ,the last condition being Left hand limit$\displaystyle \neq $Right hand limit
    Last edited by pankaj; Jun 8th 2009 at 02:20 AM.
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