<s>[INST] Proving continuity by epsilon-delta proof for a function of two variables. On account of a SE question , I raised the following question.
Let $f:D \to \mathbb R^2$ be a function in two variables. How would we go about setting up an epsilon-delta proof?
Let $f$ for example be given by $f(x,y) = (x+y)^2$ on the domain $D = \mathbb R^2$.
Edit. I am aiming at the kind of continuity that goes like the following for a one-variable function $f$. For every $\varepsilon > 0$, we have an $\delta >0$ such that $|x-a| < \delta$ implies $|f(x)-f(a)| < \varepsilon$.
 [/INST] Let's show it is continuous at $(a,b)$. Then you calculate:
$|f(x,y) - f(a,b)| = |((x-a) + (y-b) + (a+b))^2 - (a+b)^2| \leq (x-a)^2 + (y-b)^2 + 2|x-a||y-b| + 2|x-a||a+b| + 2|y-b||a+b)| < (x-a)^2 + (y-b)^2 + (x-a)^2 + (y-b)^2 + 4(|a|+|b|)\cdot \sqrt{(x-a)^2 + (y-b)^2} = 2\left((x-a)^2 + (y-b)^2\right) + 4(|a|+|b|)\cdot \sqrt{(x-a)^2 + (y-b)^2} < \left(2 + 4|a| + 4|b|\right)\cdot \sqrt{(x-a)^2 + (y-b)^2}$, if we preset $\sqrt{(x-a)^2 + (y-b)^2} < 1$. From here you can see the answer as you take:
$\delta = \text{min} \{1, \dfrac{\epsilon}{2 + 4|a|+4|b|}\}$ for a given $\epsilon > 0$.
Done.
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