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Math Help - Limits and sin

  1. #1
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    Limits and sin

    What is the Limit h->0 if

    Sin2h/sinh ???
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  2. #2
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    Quote Originally Posted by Luckyjoshua View Post
    What is the Limit h->0 if

    Sin2h/sinh ???
    for small x sin(x)~x, so

    sin(2h)/sin(h)~2h/h=2, so lim(sin(2h)/sin(h), h ->0)=2

    This is an informal argument (but is the right answer) if you need this
    done formally one good route is L'Hopitals rule. Using this we have:

    lim(sin(2h)/sin(h), h ->0)=lim([d(sin(2h)/dh]/[d(sin(h)]dh, h ->0)

    .................................=lim([2 cos(2h)]/[cos(h)], h ->0)

    .................................=lim([2 cos(2h)], h ->0)/lim([cos(h)], h ->0)=2

    RonL
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  3. #3
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    Quote Originally Posted by CaptainBlack View Post

    This is an informal argument (but is the right answer) if you need this
    done formally one good route is L'Hopitals rule.

    Watch und learn, furm die Meister:

    Express as,
    sin 2x/sin x=[(sin 2x)/x]/[sin x/x]=2[sin 2x/(2x)]/[sin x/x]
    Now,
    lim sin2x/2x=lim sin x/x =1 (limit composition rule)
    lim sin x/x=1
    Thus,
    2(1)/(1)=2
    Last edited by ThePerfectHacker; October 2nd 2006 at 06:52 PM.
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