Ok, i have this problem: Calculate: Limit x->0 of ((integral of cos t² dt between 0 and x²) / (integral of e^(-t²) dt)) So, the limit of both integrals is 0 since the interval between both integrals tends to 0. I used L'Hopital then, so: Limit x->0 F'(x)/G'(x) = (cos (x^4) * 2x)/(e^(-x²)), F(x) and G(x) being both integrals. I know that this is right, i just don't remember why F'(x) = cos (x^4) * 2x and not F'(x) = cos (x^4)... I know it have something to do with the fact that 2x is the derivative of x² but there's something missing in this explanation. Tks in advance.