L 'Hopital Rule: If h(x) = f(x)/g(x) and h(a) = 0/0 or (infinity)/(infinity) , both are differentable at a, then lim x --> a = f'(a)/g'(a) for example, if h(x) = (x-1)/logx, which is in 0/0 form at x=1, and both the top and bottom functions are differentiable, then by L'hopital rule, lim x-->1 h(x) = 1/(1/x) = 1 Got it???