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