Multivariable Delta Epsilon Proof I am trying to prove the following limits using the delta-epsilon method. Can you help me out?
$$ \lim_{(x,y)\to(a,b)}(x+y) = a+b$$
 A: Hint: For $\epsilon > 0$, choose $\delta = \epsilon / 2$. Then show that if $|x-a|  < \delta$ and $|y-b| < \delta$, then $|(x+y) - (a+b)| < \epsilon$. This should suffice for whatever choice of metric you want to use on ${\mathbb R}^2$, getting the point across without particulars about choice of metric etc. Intuitively it just shows that as $(x,y)$ gets close to $(a,b)$, you get that $x+y$ gets close to $a+b$.
A: Hint:
Since on $\mathbb R^n$ all norms are equivalent, chose the easiest: $\Vert\cdot\Vert_\infty$. This will leave you with two 1-D. $\epsilon$-$\delta$ arguments where
$$\epsilon = (\epsilon_1, \epsilon_2)$$
and you want
$$|x+y-a-b| = |(x-a)+(y-b)| \leq |x-a| + |y-b| < \epsilon_1 + \epsilon_2 = 2\Vert\epsilon\Vert_\infty$$
Get the gist? I can provide more detail if needed.

Now all that remains to prove is
$$\lim_{x\to a} x = a$$
or equivalently (by exchanging symbols)
$$\lim_{y\to b} y = b$$
For the choice of $\delta$, $\delta_\epsilon := \frac{\epsilon}{2}$ suits your needs.

Proof
Using equivalency of norms on $\mathbb R^n$, we have to prove
$$\forall\ \epsilon > 0 \quad \exists\ \delta > 0$$
such that
$$|x+y - (a+b)| < \epsilon \qquad \forall\ (x,y) \in \mathbb R^2 | \Vert (x,y) - (a,b) \Vert_\infty < \delta$$
Now for $\delta := \frac{\epsilon}{2}$ we have
$$|x+y - (a+b)| \leq |x-a| + |y-b| < \delta + \delta = 2\delta = \epsilon$$
q.e.d.

Note
Using the euclidean norm $\Vert\cdot\Vert_2$ we can chose $\delta = \frac{\epsilon}{4}$, since
$$\Vert (x,y) \Vert_2^2 = x^2 + y^2 \leq x^2 + 2|x||y| + y^2 = (|x| + |y|)^2 \leq (2\Vert(x,y)\Vert_\infty)^2$$
(actually we give away a bit here, but that's not important for the task at hand)
