# Proving limits with different epsilon-delta relationship

In the thomas textbook, the author proves the limit of 4xy^2/(x^2+y^2) as (x,y)->(0,0) by proving that the definition holds if delta equals e/4. But the definition equally holds if delta=e/5. Does that mean i can use any arbitrary constant larger than 4 for the delta epsilon ratio as the answer?

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Yes. Suppose that whenever $\sqrt{x^2+y^2}\lt \dfrac{\epsilon}{4}$, something nice happens. Let $k$ be any number $\ge 4$. Then whenever $\sqrt{x^2+y^2}\lt \dfrac{\epsilon}{k}$, the same nice thing happens.

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Yes you can. In every limit statement, strengthening the condition on delta can only make things better. With symbols, let $\delta'\leqslant\delta$ and assume that $$\text{something}\leqslant\delta\implies\text{something else}\leqslant\varepsilon.$$ Then, since $$\text{something}\leqslant\delta'\implies\text{same thing}\leqslant\delta,$$ one gets $$\text{something}\leqslant\delta'\implies\text{something else}\leqslant\varepsilon.$$

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In such problems you’re given an $\epsilon>0$, and you’re trying to find a $\delta>0$ such that whenever $\text{thing}_1$ is smaller than $\delta$, $\text{thing}_2$ is smaller than $\epsilon$. In this case you find that taking $\text{thing}_1<\frac{\epsilon}4$ is good enough to ensure that $\text{thing}_2<\epsilon$. Forcing $\text{thing}_1$ to be even smaller certainly isn’t going to hurt: if $\text{thing}_1<\frac{\epsilon}5$, then it’s certainly less than $\frac{\epsilon}4$, and that guarantees that $\text{thing}_2<\epsilon$, just as you want. If $\text{thing}_1<\frac{\epsilon}{500}$, again you can be sure that $\text{thing}_1<\frac{\epsilon}4$ and therefore that $\text{thing}_2<\epsilon$.

So yes: you can use any number smaller than $\frac{\epsilon}4$ as your $\delta$ without any harm. In particular, if $a\ge 4$, $\frac{\epsilon}a$ is a perfectly good choice for $\delta$.

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