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    lim x goes to 0()int from o to x (x^2sinx dx/x^2)
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    Let me post a picture
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    (Original post by ElliotWalton)
    ...
    Not that interesting. Just apply L'Hopitals twice?

    Alternatively can be done by just expanding \sin x using MacLaurin's then integrating the polynomial before dividing by x^2 and doing the limit
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    (Original post by RDKGames)
    Not that interesting. Just apply L'Hopitals twice?

    Alternatively can be done by just expanding \sin x using MacLaurin's then integrating the polynomial before dividing by x^2 and doing the limit
    could you please show a working and answer? I am quite confused about it.
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    (Original post by ElliotWalton)
    could you please show a working and answer? I am quite confused about it.
    Do you understand how L'Hopital's rule works?
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    (Original post by RDKGames)
    Do you understand how L'Hopital's rule works?
    absolutely i know that
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    Actually, the question isn't quite correct. You can't have x both in the integration limit and as an integration variable.
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    (Original post by ElliotWalton)
    absolutely i know that
    What Zacken said is correct and I didn't pick up on it - perhaps the question should read \displaystyle \lim_{x \rightarrow 0} \dfrac{\int_0^x t^2 \sin t .dt}{x^2}?
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    (Original post by RDKGames)
    What Zacken said is correct and I didn't pick up on it - perhaps the question should read \displaystyle \lim_{x \rightarrow 0} \dfrac{\int_0^x t^2 \sin t .dt}{x^2}?
    yes you are right. Could you please help me cope with it ?
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    (Original post by ElliotWalton)
    yes you are right. Could you please help me cope with it ?
    Then in that case you should know that applying L'Hopitals would mean you need \displaystyle \dfrac{d}{dx} \int_0^x t^2 \sin t dt, but what does the Fundamental Theorem of Calculus tells us this is?
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    (Original post by RDKGames)
    Then in that case you should know that applying L'Hopitals would mean you need \displaystyle \dfrac{d}{dx} \int_0^x t^2 \sin t dt, but what does the Fundamental Theorem of Calculus tells us this is?
    Frankly speaking,I have no idea.
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    (Original post by RDKGames)
    Not that interesting. Just apply L'Hopitals twice?
    Not sure there's even any need to apply L'Hopital - just bound the integrand by x^2.
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    (Original post by ElliotWalton)
    Frankly speaking,I have no idea.
    Quick wikipedia look at the theorem says that for F'(t) = f(t), we have \displaystyle \int_a^b f(t) .dt = F(a) - F(b) so you can just apply this.
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    (Original post by DFranklin)
    Not sure there's even any need to apply L'Hopital - just bound the integrand by x^2.
    Ah didn't notice that - not used to bounding these, but there's the third method for OP then.
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    (Original post by RDKGames)
    Quick wikipedia look at the theorem says that for F'(t) = f(t), we have \displaystyle \int_a^b f(t) .dt = F(a) - F(b) so you can just apply this.
    Thanks for your help.Let me have a try.BTW,would you mind showing a complete working of this question?
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    You don't need to calculate the whole integral to get the answer, you can just use the inequality sin t <= 1 to get that the absolute value of the integral is less than x/3 hence tends to 0 as x tends to 0.
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    (Original post by Jeremy01)
    ~snip~
    Please see the Posting Guidelines and in particular what they have to say about posting full solutions.
 
 
 
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