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April 20, 2015

April 20, 2015

Posted by **soumitra** on Friday, July 11, 2014 at 11:13am.

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

x>0

- math -
**Steve**, Friday, July 11, 2014 at 11:29amwhat 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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