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Math Help - Integral

  1. #1
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    Integral

    Hi,

    I'm a bit stuck on this question:

    Let g be differentiable in [-1,1]. g(0)=0 and g'(0)=1
    Does the limit in the attached image exist (below)? If yes, calculate it. If no, explain why not.

    I'm not really sure how to answer this at all. It came up on an exam I had recently.

    Thanks
    Attached Thumbnails Attached Thumbnails Integral-2b.jpg  
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Re: Integral

    First hint: Denote h(x)=\int_{x^3}^{x^2}\left(\int_0^tg(s^2)ds\right)  dt . Then \lim_{x\to 0}h(x)=0 , as a consequence

    \lim_{x\to 0}\frac{\sin \left(\int_{x^3}^{x^2}\left(\int_0^t g(s^2) ds\right) dt\right)}{x^8}=\lim_{x\to 0}\dfrac{h(x)}{x^8}


    P.S. Show some work and I will provide more hints if necessary.
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  3. #3
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    Re: Integral

    Quote Originally Posted by FernandoRevilla View Post
    First hint: Denote h(x)=\int_{x^3}^{x^2}\left(\int_0^tg(s^2)ds\right)  dt . Then \lim_{x\to 0}h(x)=0 , as a consequence

    \lim_{x\to 0}\frac{\sin \left(\int_{x^3}^{x^2}\left(\int_0^t g(s^2) ds\right) dt\right)}{x^8}=\lim_{x\to 0}\dfrac{h(x)}{x^8}


    P.S. Show some work and I will provide more hints if necessary.
    I don't know we got this: \lim_{x\to 0}h(x)=0.

    Also did you mean the following on the last line?
    \lim_{x\to 0}\frac{\sin h(x)}{x^8}

    Does the limit then equal infinity according to L'Hopital?

    Thanks for your help so far
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  4. #4
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    Re: Integral

    How do I know that the limit h(x) = 0 (when x approaches 0)?
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  5. #5
    MHF Contributor FernandoRevilla's Avatar
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    Re: Integral

    Quote Originally Posted by durrrrrrrr View Post
    I don't know we got this: \lim_{x\to 0}h(x)=0.
    h is continuous in [-1,1] and h(0)=0

    Also did you mean the following on the last line? \lim_{x\to 0}\frac{\sin h(x)}{x^8}
    We have \lim_{x\to 0}\frac{\sin h(x)}{h(x)}=1 so, \lim_{x\to 0}\frac{\sin h(x)}{x^8}=\lim_{x\to 0}\frac{ h(x)}{x^8}

    Does the limit then equal infinity according to L'Hopital?
    The limit presents the indetermination 0/0 . Apply L'Hopital Rule.
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  6. #6
    Grand Panjandrum
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    Re: Integral

    Quote Originally Posted by durrrrrrrr View Post
    How do I know that the limit h(x) = 0 (when x approaches 0)?
    Consider Taylor's theorem for g(x) :

    g(x)=g(0)+xg'(0)+\rho(x)

    where \lim_{x \to 0} \rho(x)/x =0

    CB
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