Actually after the first application of lhopital in part b you can kill off the 2nd limit by considering sin(sinx)/2x = (sin(sinx)/sinx)(sinx/x)(1/2) to avoid the 2nd application of the lhopital rule. Also in part a I think it is easier to apply the estimation lemma on abs(the integral) and show that it's 0 by bounding it by abs(x(1-x)) for small enough x. Your method still works perfectly for both parts though.