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find the value

lim[(cos x)^1/2 -(cos x)^1/3]/(sinx)^1/2

  • math -

    what tools do you have at your disposal? Have you learned l'Hospital's Rule yet? If so, that's what I'd use. The limit is the same as for the derivatives:

    (1/2 cos(x)^-1/2 - 1/3 cos(x)^-2/3)(-sinx) / (1/2 sin(x)^-1/2 * cosx)

    Take that sinx^-1/2 out of the bottom, and move it up top, and evaluate:

    (1/2 - 1/3)(0)*0 / (1/2 * 1) = 0

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