Help with epsilon-delta definition of a limit of a function

[azure kitsune]

Homework Statement

Let

f(x,y) = \dfrac{x^2+2xy^2+y^2}{x^2+y^2}

Prove that

\lim_{(x,y) \to (0,0)} f(x,y) = 1

Homework Equations

Definition of the limit of a function of multiple variables:

It suffices to show that for all \epsilon > 0, there exists a \delta > 0 such that for all (x,y) such that 0 < x^2 + y^2 < \delta ^2, we have |f(x,y) - 1| < \epsilon

The Attempt at a Solution

|f(x,y) - 1| = \left| \dfrac{2xy^2}{x^2+y^2} \right| = \dfrac{2|x|y^2}{x^2+y^2}

I need to bound this with an expression in terms of \delta, but I can't think of any way to do so. I noticed that the denominator is less than \delta ^2 but I can't get anywhere with that. (I end up bounding it in the wrong direction! )

Can anyone point me in the right direction? Thanks.

[Edit: Good catch Mark44!]

 

[Dick]

Try 2|x|y^2<=2|x|*(x^2+y^2). How can you bound |x| in terms of delta?

 

[Mark44]

This isn't much help, but the denominator is less than \delta ^2, which makes the overall expression larger than it would be if the denominator were equal to \delta ^2.

 

[azure kitsune]

I get that |x| = \sqrt{x^2} \leq \sqrt{x^2 + y^2} &lt; \delta so

\dfrac{2|x|y^2}{x^2+y^2} \leq \dfrac{2|x|(x^2 + y^2)}{x^2+y^2} = 2|x| &lt; 2\delta

So for any \epsilon, we can choose \delta = \epsilon/2.

Thanks Dick! That was tricky.