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March 29, 2017

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Find the limit for the given function. lim t-->0 (sin(8t))^2/t^2

  • calculus - ,

    as t-->0, we have 8t-->0, so
    lim (sin(8t))^2/t^2 =
    [lim sin(8t)/t].[limsin(8t)/t] as8t-->0
    =8.[limsin(8t)/8t].8[limsin(8t)/8t], 8t-->0
    = 8*1*8*1= 64

    you know the formula lim (sint/t)=1 as
    t-->1, aren't you?

  • calculus - ,

    yea i knw... thnxx foh ur help

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