text: l'hopital's rule? use l'hopitals rule to sokve the following  lim        ((cos(x) - 1) / x^2) x->0 read my other answer for the explanation for this one. :-)  you have zero on top and zero on bottom, so you can certainly go ahead and use l'hopitals.  derivative of top and bottom.  cos(x) -1 becomes  (-sinx) and x^2 becomes 2x.  now you have -sinx / 2x which you need to evaluate for x->0. on top you have sin(0) which is zero, and on bottom 2(0) is zero. you have 0/0 again, so l'hopital again.  -sin x becomes -cos x 2x becomes 2.  now you get -cosx / 2, evaluated for zero that's -1/2.  hope that helped, umrmathmajor