Question

find lim (c0s x -1 ) / sin x , do not use Lhopitals rule , as x ->0

Answer 1

If you don't want to use l'Hopital, you can use the Taylor expansion of sin and cos and use Big-O notation.

Answer 2

Better, multiply top and bottom by cos x + 1. Then you have
(cos^2 x - 1)/( sin x (1 + cos x) ) = - sin^2 x / ( sin x (1 + cos x) )

Cancel now sin x from top and bottom, and you have

- sin x / ( cos x + 1)

It is straight forward now to take the limit of numerator and denominator as x -> 0, as the denominator is not zero, to give 0/2 = 0.