Limit of function of two variables using Squeeze Theorem?

[cookiesyum]

Homework Statement

Determine the set of points at which the function is continuous:

f(x,y)= (x^2)(y^3)/[2x^2 + y^2] if (x,y) is not equal to (0,0)
= 1 if (x,y) is equal to (0,0)

Homework Equations

Definition of the limit of a function of two variables:

lim f(x,y) as (x,y)-->(a,b)=L if for every number E>0 there is a corresponding number D>0 such that if 0<sqrt[(x-a)^2 + (y-b)^2]< D then the absolute value of f(x,y) - L < E.

The Attempt at a Solution

f is continuous on R^2 except at possibly the origin. Since x^2< or = 2x^2 + y^2, the absolute value of (x^2)(y^3)/[2x^2 + y^2] is < or = the absolute value of y^3 (which has a limit of 0 as (x,y) approaches 0). Now how can I use this to determine the limit of f(x,y)? I'm guessing it has something to do with the Squeeze Theorem, or a direct application of the definition of the limit of a function of two variables.

 

[yyat]

Compute f(x,0) for different values of x.

 

[HallsofIvy]

Or f(0,y) for different values of y! That is, find $\lim_{x\rightarrow 0} f(x,0)$ or $\lim_{y\rightarrow 0} f(0,y)$

In order for the function to be continuous at (0,0), the limit must exist and it must be equal to the value of f(0,0).