Proving lim_(x,y)→(0,0) (x²+2xy²+y²)/(x²+y²)=1 with epsilon-delta

[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 $$|x| = \sqrt{x^2} \leq \sqrt{x^2 + y^2} < \delta$$ so

$$\dfrac{2|x|y^2}{x^2+y^2} \leq \dfrac{2|x|(x^2 + y^2)}{x^2+y^2} = 2|x| < 2\delta$$

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

Thanks Dick! That was tricky.