Proving $f$ is continuous using the $\epsilon-\delta$ definition Prove that $f(x,y) = x^2 + xy$ is continuous using the $\epsilon-\delta$ definition, at $(1,-1).$
I encountered this question in a test and I fumbled through it quite shamelessly. Here is my latest attempt.
We want to find a $\epsilon > 0$ such that when $ \rho ((x,y),(1,-1)) < \delta$ and $N \in \mathbb{N}$, we have $$|f(x,y) - f(1,-1)|< \epsilon$$ $\forall n \le N$.
$$\sqrt{(x-1)^2 + (y+1)^2} < \delta$$
$$|x-1|^2 + |y+1|^2 < \delta^2$$
So,
$$|x-1| < \delta$$ and 
$$|y+1| < \delta$$
Here is where I get reach-y.
\begin{align}
|f(x,y)-f(1,-1)| &\le& |x-1|^2 + 2 |x-1||y+1|\\
&<& \delta^2 + 2 \delta^2\\
\end{align}
In the end I take $\epsilon$ as $3\delta^2$ which seems strange to me. Is this as wrong as I think it is? I've never been able to follow this specific method in class.
 A: You have a misconception there. First you need to choose $\epsilon$ arbitrary and then you have to prove that $\delta$ with certain properties exists. $N\in\mathbb N$ doesn't have anything to do with it.
I can't see how you get from $|f(x,y)-f(1,-1)|$ to $|x-1|^2+2|x-1||y+1|$.
That's one possible way to do it:
For given $\delta$, if $\|(x,y)-(1,-1)\| < \delta$, we have, as you showed correctly: $|x-1|<\delta$ and $|y+1|<\delta$. Thus, it follows:
$|f(x,y)-f(1,-1)| = |f(x,y)-0| = |x^2+xy| = |x|\cdot |x+y|$
$ = |x-1+1|\cdot |x-1+y+1| \leq (|x-1|+1)\cdot (|x-1|+|y+1|) < (\delta+1)\cdot (2\delta) = 2(\delta^2+\delta).$
So we now need to show that for given $\epsilon$ it is possible to chose $\delta>0$ s.t. $2\delta ^2 +2\delta \leq \epsilon$.
Edit: If we assume $\delta<1$ we obtain $\delta^2<\delta$. Hence $2\delta^2+2\delta < 4\delta$. If we now choose $\delta := \min\{\frac{\epsilon}{4},1\}$ for given $\epsilon$, we have $|f(x,y)-f(1,-1)| < 4\delta = \epsilon$ if $\|(x,y)-(1,-1)\|<\delta$.
(Thanks for the hint, marty cohen)
A: If you have $\sqrt{(x-1)^2 + (y+1)^2} < \delta$, then in particular you have $|x-1|<\delta$ and $|y+1|<\delta$, so we can work with there two last inequalities. Note that $f(1,-1) = 0$. Suppose that $\delta < 1$, also. Then $|x|,|y| < 2$. Then: $$\begin{align}|x^2+xy| &= |x||x+y|\\ &\leq 2|x+y| \\ &= 2|x-1+1+y| \\ &\leq 2(|x-1|+|y+1|) \\ &\leq 2(2\delta) =4\delta. \end{align}$$
This suggests that $\delta = \min\{1, \epsilon/4\}$ will work. No need to solve second degree equations, this is not efficient at all. A rigorous proof would be: 


*

*Let $\epsilon > 0$ and choose $\delta = \min\{1,\epsilon/4\} > 0$. If $(x,y)$ verify $\sqrt{(x-1)^2+(y+1)^2}<\delta$, then we have $|x-1|,|y+1|< \delta$. Since $\delta < 1$, this gives $|x|<2$. So: $$|f(x,y)-f(1,-1)| = \cdots \leq \cdots \leq \cdots < \epsilon.$$

