Proving by definition $\lim_{(x,y)\to(-1,8)} xy = -8$

Question

I need to prove by definition the following limit:

$$\lim_{(x,y)\to(-1,8)} xy = -8$$

I've reached a point where I don't know how to proceed.

So given an $\epsilon > 0$ I need to find a $\delta>0$ such that $|f(x,y)+8|<\epsilon$ when $\left\lVert (x,y)-(-1,8)\right\rVert<\delta$.

Knowing the distance from $(x,y)$ to the point $(-1,8)$ is less than delta I also know that $|x+1|<\delta$ and $|y-8|<\delta$. So I try to use that in the inequation involving $\epsilon$.

$|xy + 8|<\epsilon$

$|xy + 8+1-1+x-x+y-y|<\epsilon$

$|2(x+1)+(y-8)+y(x+1)| \leq 2|x+1| + |y-8| + |y| |x+1|<\epsilon$

$|2(x+1)+(y-8)+y(x+1)| \leq 2|x+1| + |y-8| + |y| |x+1|< 3 \delta + |y| \delta < \epsilon$

So pretty much I don't know what to do with the $|y| \delta$ in $3 \delta + |y| \delta < \epsilon$

Any hints? Thanks!!

Answer 1

You can first require that $\delta < 1$. Then $|y-8| < \delta$ implies $|y| < 9$, leading to $3\delta + 9\delta < \epsilon$. This will hold if $\delta < \frac{\epsilon}{12}$, so you can make two conditions on $\delta$, i.e. $\delta < \min \{1,\frac{\epsilon}{12} \}$.