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Math Help - first post

  1. #1
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    first post

    Hi, my name is Mike. I'm a student of Cal State Fullerton. I have this problem with L'hospital rule:

    the Problem: find values of a and b that satisfy the following

    lim (sin x/x^3 + a + b/x^2) = 0
    x->0

    I tried to combined the whole equation into one using common denominator but it's doesn't seem right at all because I have 2 unknown variables.
    Last edited by ninedragon; November 7th 2012 at 08:37 PM.
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  2. #2
    MHF Contributor MarkFL's Avatar
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    St. Augustine, FL.
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    Re: first post

    Hello and welcome, Mike!

    I recommend in the future posting your problems in the appropriate forums, as they will receive more and prompt attention that way.

    I think your first move is a good one, i.e., to write:

    \lim_{x\to0^{+}}\frac{\sin(x)+ax^3+bx}{x^3}=0

    We have the indeterminate form 0/0, so application of L'H˘pital's rule gives:

    \lim_{x\to0^{+}}\frac{\cos(x)+3ax^2+b}{3x^2}=0

    In order for this limit to be bounded, we require the numerator to tend to 0, so we need b=-1, and we may write:

    \lim_{x\to0^{+}}\frac{\cos(x)+3ax^2-1}{3x^2}=0

    We have again the indeterminate form 0/0, so application of L'H˘pital's rule gives:

    \lim_{x\to0^{+}}\frac{-\sin(x)+6ax}{6x}=0

    We have again the indeterminate form 0/0, so application of L'H˘pital's rule gives:

    \lim_{x\to0^{+}}\frac{-\cos(x)+6a}{6}=0

    Can you now finish?
    Last edited by MarkFL; November 7th 2012 at 08:47 PM.
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