f(x) f'(0) lim ------ = ------ x->0 g(x) g'(0)evaluating rational functions at certain values can give problems, such as what is the result of dividing 0 by 0 in this:

sin(x) lim ------- (actually it is = 1!) x->0 xL'Hopital's rule is "The limit of a rational function can be found by replacing the top and bottom function by their derivatives." [1]For the sin(x)/x case, d/dx(sin(x))=cos(x); d/dx(x)=1. So the limit is cos(0)/1 = 1. Other examples:

sin(k.x) lim ------ = k (= k.cos(0)) x->0 x (1-cos(x)^2) 1 { sin(0) cos(0) } lim ------------ = --- {= ------ = ------ } x->0 x^2 2 { 2.[x=0] 2 } sin(x) lim ------ = 1/k (= cos(0)/k) x->0 k.xIF the ratio of derivatives is also 0/0, use the second derivative and so on as in the second case above.

*[Simplified proof of L'Hôpital: if the slope of the top part of fraction is N times the slope of the bottom part then the ratio of the values near the zero value is N (go to dx, top part has value N.dx, bottom value is dx).]*The L'Hôpital rule can also be applied at infinity, where the numerical method should have a problem:

exp(x) lim ------ = Inf (= exp(Inf)/1 since d/dx(exp(x))=exp(x) & d/dx(x)=1) x->Inf xApplication of L'Hopital rule may save considerable effort in coding a means of evaluating the limit.