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lim (4sin2x - 3x cos5x)/(3x/2 +(x^2)cscx )
x->0

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mo

  1. L'Hoptial's rule applies to this.

    I get for the numerator..

    8cos2x - 3cos5x - 15x sin5x

    and the denominator...
    3/2 + 2x csc x + x^2 d cscx/dx

    so the limit looks to be
    5/ (3/2) = 10/3

    check me.

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    bobpursley

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