I need to find the limit as x,y->0 of (x^2+y^2)(ln(x^2+2y^2)) anylitically. since the limit of this is 0(-infinity) which is indeterminant .. I tried to approach it along the line y=x which gives: lim(x->0) of [2x^2*ln(3x^2)]. Again, that gives 0(-infinity). Now, I haven't done calculus in 6 months .. I know I need to get it to a quotient indeterminant form in order to be able to use l'hopital's rule .. I tried: lim (x->0) of [2x^2/(ln(3x^2)^-1] but that gives 0/(1/(-infinity) .. that's indeterminant, so I can take the derivitive, and get: lim (x->0) of [4x/(-1(ln(3x^2)^-2)(1/(3x^2))(6x)] :yuck: which is lim (x->0) of [(4x*3x^2)/(-6xln(3x^2)^-2)] which is 0/0 .. I don't think this is getting me anywhere. Any ideas on how to make this work? :yuck: Thanks!