Seems simple enough, but my notes do something I don't understand.
Question is:
prove lim xy = 0
(x,y) -> (0,0)
The notes say that
0 <| (x,y) - (0,0) | < d
implies 0 < |(x, y)| < d
step 3: implies 0 < square root (x^2 + y^2) < d
Then does:
0 <= (x+y)^2 = x^2 + 2xy + y^2
implies 2xy <= x^2 + y^2
implies 2|xy| <= x^2 + y^2
implies |xy| <= x^2 + y^2 / 2 < d/2
and chooses epsilon = d^2 /2 which impliesd^2 = epsilon / 2 which implies to pick d = square root of (epsilon/2).
Im confused on step 3. Why did they take the square root of (x^2 + y^2)? How does step 2 imply step 3?
And why was step 3 followed with
0 <= (x+y)^2 = x^2 + 2xy + y^2
Where did that come from?
I was taught to go from |xy - 0| < epsilon and try to make |xy - 0| to something like |(x,y) - (0,0)|
They took the square root of $x^2+y^2$ because $\bigl\lVert(x,y)\bigr\rVert=\sqrt{x^2+y^2}$.
Step 2 implies step 3 because $\bigl\lVert(x,y)-(0,0)\bigr\rVert=\bigl\lVert(x,y)\bigr\rVert$.
Finally, since$$\lvert xy\rvert\leqslant\frac{\sqrt{x^2+y^2}}2,\tag1$$then, for every $\varepsilon>0$, if you take $\delta$ such that $\delta=2\varepsilon$, it follows from $(1)$ that$\sqrt{x^2+y^2}<\delta\implies\lvert xy\rvert<\varepsilon$. Yes, I know that this is not the same $\delta$ as the one that you mentioned, but I suppose that what matters to you is that some $\delta$ will do.