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Math Help - limit of integral

  1. #1
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    limit of integral

    Evaluate attachment

    (I uploaded the equation rather than typing it out as my previous post).

    Thanks.
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by arianas89 View Post
    Evaluate attachment

    (I uploaded the equation rather than typing it out as my previous post).

    Thanks.
    \lim_{x\to{0}}\frac{1}{x^3}\int_0^x \sin(t^3)\,dt=\lim_{x\to{0}}\frac{\int_0^x \sin(t^3)\,dt}{x^3}

    Applying L'Hopital's Rule:

    \lim_{x\to{0}}\frac{\int_0^x \sin(t^3)\,dt}{x^3}=\lim_{x\to{0}}\frac{\sin(x^3)}  {3x^2}

    Apply L'Hopital's Rule again...

    \lim_{x\to{0}}\frac{\sin(x^3)}{3x^2}=\lim_{x\to{0}  }\frac{3x^2\cos(x^3)}{6x}=\lim_{x\to{0}}\frac{1}{2  }x\cos(x^3)=\color{red}\boxed{0}

    I hope this clarifies things!

    --Chris

    P.S. Anyone who reads this, is there another way to do this w/o L'Hopital's Rule?? (ahemm....Mathstud...)
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  3. #3
    Moo
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    Hello,
    P.S. Anyone who reads this, is there another way to do this w/o L'Hopital's Rule??
    I've never done it this way, so it's experimental ^^

    When x approaches 0, one can say that t and hence t^3 approaches 0 because it's restricted below by 0 and above by x.

    So \sin(t^3) \sim t^3

    Therefore \lim_{x \to 0} \frac{\int_0^x \sin(t^3) \ dt}{x^3}=\lim_{x \to 0} \frac{\int_0^x t^3 \ dt}{x^3}=\lim_{x \to 0} \frac{x^4}{4x^3}=\lim_{x \to 0} \frac x4=0
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Moo View Post
    Hello,

    I've never done it this way, so it's experimental ^^

    When x approaches 0, one can say that t and hence t^3 approaches 0 because it's restricted below by 0 and above by x.

    So \sin(t^3) \sim t^3

    Therefore \lim_{x \to 0} \frac{\int_0^x \sin(t^3) \ dt}{x^3}=\lim_{x \to 0} \frac{\int_0^x t^3 \ dt}{x^3}=\lim_{x \to 0} \frac{x^4}{4x^3}=\lim_{x \to 0} \frac x4=0
    (OP) you can do this for as you noted we are considering \sin(x^3) in the subset of the reals described by the limits of integration and the radius of this region centered at zero is approaching zero. Therefore you
    may apply asymptotic equivalences.
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