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What is 0 multiplied by infinity in limits?

  1. Oct 15, 2013 #1
    If you have f(x) = 1/x and g(a) = (cosx - 1)/x and then y = [limx→0 f(x)][limx→0g(x)], the two individual limits equal 0 and infinity, respectively. Since these are limits and only approach these values, would the multiplication of the two limits equal 0, infinity or something else?
     
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  3. Oct 15, 2013 #2

    Office_Shredder

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    In this case, [tex]y= \lim_{x\to 0} f(x) \lim_{x\to 0} g(x) [/tex] is a meaningless expression. You CAN calculate
    [tex] \lim_{x\to 0} f(x) g(x) [/tex]
    but you cannot split it into two limits and give the new expression any meaning.
     
  4. Oct 15, 2013 #3

    Mark44

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    If you take a limit and get 0 * ∞, it means that you're not done yet. The indeterminate form [0 * ∞] is one of several indeterminate forms that can arise when you're taking limits. One thing that is often done is to rewrite the expression so that the limit becomes [0/0] or [∞/∞], either of which might be amenable to evaluation using L'Hopital's Rule.

    This wiki article has a table of indeterminate forms - http://en.wikipedia.org/wiki/Indeterminate_form#List_of_indeterminate_forms
     
  5. Oct 15, 2013 #4

    SteamKing

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    The limit of f(x) = 1/x as x approaches 0 is not zero, it is infinity.
     
  6. Oct 15, 2013 #5

    Mark44

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    You're sort of half correct.
    $$\lim_{x \to 0^+}\frac 1 x = ∞$$
    $$\lim_{x \to 0^-}\frac 1 x = -∞$$
    Since the one-sided limits aren't equal, the two-sided limit doesn't exist.
     
  7. Oct 15, 2013 #6
    Sorry for my poor phrasing.

    How about in this new example.

    y = [limx→0^+ 1/x][limz→0 (1 - cosz)/z]

    Would it be possible to evaluate this limit since one limit approaches infinity while the other approaches 0, and they are different variables in this case.
     
    Last edited: Oct 15, 2013
  8. Oct 15, 2013 #7

    Mark44

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    No, you're right back to what I was talking about in post #3, and what you had in your first post. The fact that you have different variables does not change things.
     
  9. Oct 15, 2013 #8

    Mark44

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    How did this limit come up? What's the problem you're trying to solve?
     
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