I am trying to jump doing $\epsilon-\delta$ proofs of the limit of functions of one variable (i.e. $f:\mathbb{R} \to \mathbb{R}$), to functions of two variables (i.e. $g: \mathbb{R}^2 \to \mathbb{R}$) and I am struggling since some of the methods I learned for one variable do not immediately carry for two variables. Unfortunately, I find that my books are too eager to jump to nice theorems to calculate limits (limit of a sum is equal to the sum of the limits, etc.) but I insist on being able to find limits to the most elementary functions like $x+y,xy, x^2 + y^2, \frac{x}{y}, \frac{1}{x+y}$ using the $\epsilon-\delta$ definition (without more advanced notions like the topology of $\mathbb{R^n}$) of the limit until I gain some intuition to move on and prove more general theorems using this technique. In this post, I will show you one case I think I was able to work out correctly, and another where I am struggling.
1) Perharps the simplest case I could think was to show that
$$\lim_{(x,y)\to(a,b)}(x+y) = a+b$$.
In this case, the definition says that $f(x,y)=(x+y) \to (a+b)$ when $(x,y) \to (a,b)$ if $\forall \epsilon > 0$, $\exists \delta > 0$ such that for all $(x,y)$ in the domain of $f$, whenever $0<\sqrt{(x-a)^2+(y-b)^2}<\delta$, we have $|f(x,y) - (a+b)|<\epsilon$. For this, all I had to do was to realize that $\sqrt{(x-a)^2+(y-b)^2}<\delta$ implies that $|x-a|<\delta$ and $|y-b|<\delta$, therefore $|x + y - a -b| \le |x-a| + |y-b| < 2\delta$, therefore I should choose $\delta = \epsilon/2$. Is this line or reasoning correct?
2) Now, consider
$$ \lim_{(x,y)\to(a,b)}xy = ab $$
working backwards, I want
$$ \begin{align} |xy - ab| &< \epsilon\\ |(x-a+a)(y-b+b) -ab| &<\epsilon \\ |(x-a)(y-b) +b(x-a) +a(y-b)| &< \epsilon \qquad (*) \\ &\vdots\\ \sqrt{(x-a)^2+(y-b)^2}&<\delta \end{align} $$
(*) I got this idea from here. But that's it I am afraid. I don't know how to take it from there. I would appreciate any hint or guidance. Sorry if this post is too long. I wanted to provide context and background and would be happy to edit or split questions. Thank you in advance.
Note:
$ |x-a| \le \sqrt{(x-a)^2+(y-b)^2}$, and
$|y-b| \le \sqrt{(x-a)^2+(y-b)^2}.$
Let $\epsilon >0$ be given.
Choose
$\delta < \min(1, \dfrac{\epsilon}{1+|a|+|b|})$.
Then
$\sqrt{(x-a)^2+(y-b)^2} \lt \delta$ implies
$|xy-ab| \le$
$ |x-a||y-b| +$
$|b||x-a| +|a||y-b| \lt$
$(\delta)^2+(|b|+|a|)\delta \lt $
$(1+|a|+|b|)\delta \lt \epsilon.$
Note: $(\delta)^2 \lt \delta$ for
$ \delta =\min (1,\dfrac{\epsilon}{1+|a|+|b|})$.