Proving a limit involving multiple variables using epsilon-delta

Question

I'm having some issues with $\varepsilon$-$\delta$ proofs of limits with more than one variable. I understand the $\varepsilon$-$\delta$ definition of a limit, but I don't know how to deal with multiple variables.

Here's a simple example: prove the following limit, if it exists, using the epsilon-delta definition: $$ \lim_{(x,y)\to (1,2)}\frac{x^2}{x+y} = a \iff \forall(\varepsilon>0)\, \exists(\delta) \left[ |(x,y)-(1,2)|<\delta \implies \left|\frac{x^2}{x+y}-a\right|<\varepsilon \right] $$

Obviously this limit is $1/3$, but how do I prove it?

Answer 1

Let $x= 1 +t\;\;y=2+s$ Let $s^2 +t^2<\delta^2$ $$\dfrac{x^2}{x+y}=\dfrac{1}{3+t+s} +\dfrac{2t+t^2}{3+t+s}$$ $$|\dfrac{2t+t^2}{3+t+s}| \leq |\dfrac{2t+t^2}{3-2\delta}|\leq3\dfrac{\delta}{3-2\delta}\leq3\delta$$ If $\delta<1.$

$$|\dfrac{1}{3+t+s}-\frac13|=|\dfrac{t+s}{3(3+t+s)}|\leq\dfrac{2\delta}{3(3-2\delta)}\leq\dfrac{2\delta}{3}$$ Using this you can complete the proof.

Answer 2

Actually, It's quite obvious. Whenever we have multiple variables involved, look for the interval that the variables are in, and we'll able to find a bound (upper or lower) for the variables.

For example, in your example, the interval for (x,y) is (1,2). Thus, I claim x < 1 and y < 2 respectively, and note the inequality are strict, since this interval is not closed. So then, when I plug these bounds into the last part of your implication, I get 1 / (1+2) = 1/3 < $\epsilon$

I hope this helps.