# Limits of Multivariable equations

1. Oct 24, 2005

### cappygal

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!

2. Oct 25, 2005

### Tom Mattson

Staff Emeritus
Instead of changing $\ln\left(3x^2\right)$ to $\left(\ln\left(3x^2\right)\right)^{-1}$, try changing $2x^2$ to $\left(2x^2\right)^{-1}$.