many value of delta corresponding to single epsilon?

Question

I have to prove that limiting value of $f(x,y)=xy^2/(x^2+y^2)$ is zero when $(x,y) \to (0,0)$ using epsilon delta relation.

My solution is $|xy^2/(x^2+y^2)|< |x|< \delta$ as $\sqrt{x^2+y^2}<\delta$ so here I got $\varepsilon=\delta$. Now I can also take $\varepsilon= n \times \delta$ where $n>1$

In this way I have many value of $\delta$ corresponding to one value of $\varepsilon$. Is that correct ?

Answer 1

Right, $0\leq\left|\frac{xy^2}{x^2+y^2}\right|\leq|x|.$

The limite 0 follows immediately from the squeeze theorem. (are you familiar with?)

Answer 2

The number $\varepsilon$ is fixed from the start. Therefore, you can't take $\varepsilon=n\times\delta$.

On the other hand, if $\delta=\varepsilon$ works, then $\delta=\frac\varepsilon2,\delta=\frac\varepsilon3,\ldots$ will also work. There's no problem there.