Epsilon delta proof; constraining delta.

Question

I've been trying to do some $\epsilon -\delta$ proofs, but I keep running into problems regarding certain steps. Namely, I can bring the proof to a point that is almost complete, but to complete it I would need to place a constraint on $\delta$ to make the next inequality true (such as $x^2+y^2≤\sqrt{x^2+y^2}$ provided $\sqrt{x^2+y^2}≤1$).

If I had $$\lim_{(x,y)\to(0,0)}f(x,y)$$ Is it fine to constrain my $\delta$ (such as $0<\sqrt{x^2+y^2}<\delta<1$) because my limit is within the disc $\sqrt{x^2+y^2}<1$?

If I come to a point where I would need to do that, have I done something wrong?

Is there some extra step that I should make that I am missing?

Answer 1

The short answer would be: No, you have done nothing wrong. And here's why:

The definition of limit says that for every $\varepsilon > 0$, there exists a $\delta > 0$, such that... etc. If you place any (extra) constraints on what $\delta$ may be, then nothing goes wrong. In your example you say you choose $0 < \delta < 1$ and then you get what you want. That's fine because that means a suitable $\delta > 0$ definitely exists and that's really all we need.

So yes, you may place any constraint on $\delta>0$ that you want.

N.B. Just make sure that you keep $\delta > 0$. So the constraint $\delta \le 0$, would not be fine for instance. This goes without saying, however.