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How to show lim sin(1/x) DNE as x->0

  1. May 5, 2005 #1
    How to show lim sin(1/x) DNE as x-->0


    I am trying to see how lim sin(1/x) does not exist as x-->0. It is obvious from a graph.

    I think one way to do this is to pick two sequences converging to 0 and show that the limit of these sequences do not equal each other. For example, I can pick a sequence where sin gives me +1 and another one where sin gives me -1. I think this would work. Can someone confirm that this strategy is sound?

    Also, I would also like to see how to show this using epsilon-delta argument at well.

    Thanks in advance.
  2. jcsd
  3. May 5, 2005 #2


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    Yes, that will work. If the limit of a continuous function exists at a: that is, if
    [tex]lim_{x->a} f(x)[/tex] exists, then for any sequence {xn}, [tex]lim_{n->\inf }f(x_n)[/tex] exists and is the same. If, for any such sequence the limit does not exist, or if two such sequences do not have the same limit, then the limit itself does not exist.
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