Prove $\lim_{(x,y) \to (-1,8)} xy = -8$ using only the definition.

Question

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?

Answer 1

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 $\epsilon$

Answer 2

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.