Problem with two variable limit where $\lim\limits_{(x,y) \to (-1, 8)} xy = -8$ using only definition So we just started with two-variable limits. The definition is quite straight forward though my head is still giving it a few spins. I thought doing a couple of examples would help me. Come the second problem though I've hit a wall. 
Consider this.
$\lim_{(x,y) \to (-1,8) } xy = -8$
So I start with the definition. For every epsilon bigger than zero, there has to be a delta bigger than zero such that $\ 0 < ||(x,y) - (-1,8)||_2 < \delta \implies |xy - (-8)|<\epsilon$
But hold and behold. I don't know where to follow from this. In class, the examples given always started trying to use the left hand side part of the inequality of $ \epsilon$ to try and find a bound for $ |xy - (-8) |$. Generally using some absolute value properties and know inequalities like $\ ||X||_ \infty \leq ||X||_2 \leq \sqrt{n} ||X||_2,$ n=elements in X.
But this fails. I don't know what to do with $\ |xy- (-8) | $. 
I can see this will be the same as $\ |x||y|+8 $. But how can I use my hypothesis or some known inequality here to prove the limit?
 A: The idea is that you want to somehow relate the quantities 
$\vert (x,y)-(-1,8)\vert$ and $\vert xy-(-8)\vert $ 
in such a way that the latter is small whenever the former is.
With this in mind, write 
$\vert xy+8\vert=\vert xy+y-y+8\vert =\vert y(x+1)+(-1)(y-8)\vert \leq \vert y(x+1)\vert +\vert y-8\vert $ 
and the result follows. More precisely: 
let $\epsilon >0$ be given.
Take $\delta _{1}=\min \left \{ 1,\epsilon  \right \}$ and $\delta _{2}=\frac{\epsilon }{9}$ and note that if $\vert y-8\vert <\delta _{1}$ then $y<9$.
If we take $\delta =\min \left \{ \delta _{1},\delta _{2} \right \}$ then 
$\vert x+1\vert, \vert y-8\vert <\delta \Rightarrow \vert xy+8\vert\leq\vert y(x+1)\vert +\vert y-8\vert =9\frac{\epsilon }{9}+\epsilon =2\epsilon$.
To finish, simply note that if $(x,y)\in B_{\delta }(-1,8)$, then $\vert x+1\vert $ and  $\vert y-8\vert <\delta $
A: use method like $$\begin{align}|f(x)g(x)-LM|&=|f(x)g(x)-Lg(x)+Lg(x)-LM|\\&=|g(x)(f(x)-L)+L(g(x)-M)|\\&\le|g(x)||f(x)-L|+|L||g(x)-M|\end{align}$$
