Posted by soumitra on Friday, July 11, 2014 at 11:13am.
find the value
lim[(cos x)^1/2 (cos x)^1/3]/(sinx)^1/2
x>0

math  Steve, Friday, July 11, 2014 at 11:29am
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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