Posted by soumitra on .
find the value
lim[(cos x)^1/2 (cos x)^1/3]/(sinx)^1/2
x>0

math 
Steve,
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