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Math Help - L'hospital rule

  1. #1
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    L'hospital rule

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

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

    I did combine the whole function using common denominators

    some one post this
    Quote Originally Posted by MarkFL2 View Post

    \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?
    But why b= -1? does it has to be that exact number?
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: L'hospital rule

    If b is any other value, then the limit is no longer bounded. Do you see why?
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  3. #3
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    Re: L'hospital rule

    ok, i see it now
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    Re: L'hospital rule

    another problem, probably still relate to L'hospital rule: lim x->0+ x^(x^2)
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  5. #5
    MHF Contributor MarkFL's Avatar
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    Re: L'hospital rule

    I have another recommendation for getting the best help here, but only for your next question, no need to do this for this question:

    Create a new topic for a new question. It is best to only post follow-up questions to the original problem in the same topic. I only mention these recommendations so that you will get the most prompt help.

    We are given (assuming the limit exists):

    \lim_{x\to0^{+}}x^{x^2}=L

    Now, take the natural log of both sides, and use \ln\left(\lim_{x\to c}f(x) \right)=\lim_{x\to c}\ln\left(f(x) \right), and using the properties of logs, see if you can get one of the forms for which L'H˘pital's rule may be applied.
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