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Math Help - limit of trig function

  1. #1
    DBA
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    limit of trig function

    lim t->0 tan6t/sin2t

    I did

    lim t->0 tan6t/sin2t=

    lim t->0 (6sin6t/6t * 1/cos6t * 2t/2sin2t)=

    6 * lim t->0 (sin6t/6t ) * lim t->0 (1/cos6t) * 1/2 * lim t->0 (2t/sin2t)=

    6 * 1 * lim t->0 (1/cos6t) * 1/2 * 1


    So I stick with the middle term. In the solution they have
    lim t->0 (1/cos6t) = 1/1
    but I do not know how they get that.

    I have the formulas:

    lim x->0 sinx/x = 1
    lim x->0 (cosx-1)/x = 0

    d/dx sinx = cosx
    d/dx cosx = -sinx

    Can someone help me with that?
    Thanks
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  2. #2
    DBA
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    Is it possible that I can use direct substitution here, since cos6t cannot get zero?
    cos(0)=1

    so
    lim t->0 (1/cos6t) = 1/ cos(0) = 1/1 = 1

    Then
    lim t->0 tan6t/sin2t = 6 * 1 * 1 * 1/2 * 1 = 3

    I that right?
    Thanks
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  3. #3
    Flow Master
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    Quote Originally Posted by DBA View Post
    Is it possible that I can use direct substitution here, since cos6t cannot get zero?
    cos(0)=1

    so
    lim t->0 (1/cos6t) = 1/ cos(0) = 1/1 = 1 Mr F says: Yes.

    Then
    lim t->0 tan6t/sin2t = 6 * 1 * 1 * 1/2 * 1 = 3

    I that right?
    Thanks
    Yes.
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