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Homework Help: Limit question

  1. Jan 24, 2005 #1
    Either I can't find the correct la text or I'm doing it wrong but here goes:

    for a limit where x-->0 for sin 1/x

    Am I just supposed to recognize that as x goes to zero from the left that it goes to negative infinity and as x goes to zero from the right it goes to positive infinity? What else could I deduce from this?
     
  2. jcsd
  3. Jan 24, 2005 #2
    the problem is same as
    [tex] \lim_{x \rightarrow \infty} sin(x) [/tex]
    the limit will oscillate from -1 to 1
    (not all limit is define)
     
  4. Jan 24, 2005 #3
    Thank you.
     
  5. Jan 25, 2005 #4
    For [tex] \lim_{x \rightarrow 0} sin \frac{1}{x} [/tex] could you please explain how it is the same as[tex] \lim_{x \rightarrow \infty} sin(x) [/tex] ? I understand that not all limits can be defined but is it oscillating between -1 and 1 because it can't be defined?
     
  6. Jan 25, 2005 #5

    dextercioby

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    You need to specify how that limit goes to zero.Either "goes down or up"...

    [tex] \lim_{x\nearrow 0} [/tex] or [tex] \lim_{x\searrow 0} [/tex]

    Daniel.
     
  7. Jan 25, 2005 #6
    [tex] \lim_{x \rightarrow 0^+} sin1/x = \lim_{u \rightarrow {+ \infty}} sin u [/tex]
    [tex] \lim_{x \rightarrow 0^-} sin1/x = \lim_{u \rightarrow {- \infty}} sin u [/tex]

    substitude u=1/x and you will see why
    sory for the sloppy notation in my first post
     
  8. Jan 25, 2005 #7

    HallsofIvy

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    However, since none of those limits exist, the distinction is moot.
     
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