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?

With the inequality 
$$ |xy+8| \leq |s(t-1)| + 8|t| $$
you are on the right way. The idea now is that $s$ and $t$ should be very small since the limit takes $x$ to $-1$ and $y$ to $8$. Can you bound $|s|$ and $|t|$ in terms of $\delta$? After this you should be able to proceed to bound $|xy+8|$ in terms of $\delta$. Then you can decide how you want to choose $\delta$ depending on $\epsilon$
