Prove $\lim_{(x,y) \to (-1,8)} xy = -8$ using only the definition. I'm having problems trying to prove this limit using only the definition with delta and epsilon.
I need to see that:
$$ \forall\epsilon \  \exists \delta \ : \sqrt{(x+1)^2+(y-8)^2} < \delta \Rightarrow |xy+8|<\epsilon$$
I want to make $|xy+8|$ look like the first half. I start by replacing $t=x+1$, so $x=t-1$ and $s=y-8$, so $y=s+8$. So I get:
$$ |(t-1)(s+8)+8| = |ts+8t-s-8+8| = |s(t-1)+8t| \leq |s(t-1)|+8|t|$$
Here I'm stuck, if I use that $\delta < 1$ then I can say that $t-1 > 0$ and since $|s(t-1)| = |s||t-1|$ then $|s||t-1| = |s|t-|s|$, which means I get:
$$ |(t-1)(s+8)+8| \leq |s|t-|s|+8|t| \leq |s|t+8|t| $$
I think I need to bound it by $||(s,t)||$, so I can add/substract and simplify, but I can't see how to do it. Any tips you can give me?

For a slightly different approach consider the following inequality. \begin{align} |xy + 8| &= |(x+1)(y-8) \, \, + 8x - y + 16 | 
\\ &= |(x+1)(y-8)\, \, + 8(x+1) - (y-8)|  
\\&\leq  |(x+1)(y-8)|+ 8|x+1| + |y-8| 
\\&= |x+1|\cdot|y-8|\, + 8|x+1| + |y-8|\end{align}
If you want to use the euclidean norm, then note that using the above inequality we have
\begin{align} \| (x,y) - (-1,8) \| < \delta_0 &\Rightarrow |x+1| < \delta_0 \text{ and } |y-8| < \delta_0 \\&\Rightarrow  | xy + 8 | < \delta_0^2 + 8\delta_0 + \delta_0.\end{align}
Therefore given $\epsilon > 0$, any $\delta$ such that $0 < \delta^2 + 8\delta + \delta < \epsilon$ will work. 
