# Proof that $\lim_{(x,y)\rightarrow(0,0)} f(x,y) = 2x + 2y = 0$ exists

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.

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.