TITLE: Proof that $\lim_{(x,y)\rightarrow(0,0)} f(x,y) = 2x + 2y = 0$ exists
QUESTION [5 upvotes]: I'm trying to prove via $\delta$-$\epsilon$ argument that:
$$\lim_{(x,y)\rightarrow(0,0)} f(x,y) = 2x + 2y = 0$$
What i do is, given that:
$\lvert x \rvert \leq \sqrt{x^2 + y^2} < \delta$ and $\lvert y \rvert \leq \sqrt{x^2 + y^2} < \delta$
I sum the two inequalities, giving:
$$\lvert x\rvert + \lvert y \rvert < 2\delta$$
Multiplying both sides by 2, and using the triangular inequality:
$$2\lvert x + y\rvert \leq 2\lvert x \rvert + 2\lvert y \rvert < 4\delta$$
So using $\delta = \frac{\epsilon}{4}$ the proof is complete.
Did i made any mistakes?

REPLY [5 votes]: This looks fine. For semantic purposes, you might wish to begin your proof with "Let $\epsilon>0$ be given." Then simply fix $\delta=\frac{\epsilon}{4}$ and show that 
$$ \lvert f(x,y)\rvert<\epsilon$$
for $\lvert (x,y)\rvert<\delta.$ This is just being nitpicky – of course. Your work is sound.