Multivariable $\epsilon-\delta$ proof verification

Question

Prove that $\lim\limits_{(x,y) \to (1,1)} xy=1$

Of course, I am aware that this is "obvious", but I want to add some rigor to it. When I searched around for multivariable limits using $\epsilon-\delta$, most of the examples had $(x,y) \rightarrow (0,0)$, but in this case I have $x$ and $y$ approaching something else.

$(x,y) \rightarrow (1,1) \Leftrightarrow \lvert\lvert (x,y)-(1,1)\lvert\lvert \rightarrow 0$ which can be written as $0 < \sqrt {(x-1)^2+(y-1)^2} < \delta$ for some arbitrarily small $\delta >0$.

Goal: show that $\forall$ $\epsilon>0$ $\exists$ $\delta>0$ such that

$0 < \sqrt {(x-1)^2+(y-1)^2}<\delta\Rightarrow0<|xy-1|<\epsilon$

Proof:

If $0 < \sqrt {(x-1)^2+(y-1)^2}<\delta$, then

$|xy-1|=|xy-x-y+1+x+y-2|=|(x-1)(y-1)+(x-1)+(y-1)|$

$\le|(x-1)(y-1)|+|x-1| +|y-1|=|x-1||y-1|+|x-1|+|y-1|$

$=(\sqrt{(x-1)^2})(\sqrt{(y-1)^2})+\sqrt{(x-1)^2}+\sqrt{(y-1)^2}$

$\le(\sqrt{(x-1)^2+(y-1)^2})^2+2\sqrt{(x-1)^2+(y-1)^2}<\delta^2+2\delta$

If $\delta=$$\epsilon \over4$, then $\delta^2+2\delta=\frac{\epsilon^2}{16}+\frac{8\epsilon}{16}=\frac{\epsilon^2+8\epsilon}{16}$

Now, $\frac{\epsilon^2+8\epsilon}{16}<\epsilon \Leftrightarrow \epsilon^2+8\epsilon<16\epsilon\Leftrightarrow\epsilon(\epsilon-8)<0$

Since $\epsilon>0$, we have $\epsilon(\epsilon-8)<0$ if and only if $\epsilon<8$

So if we have $\epsilon<8$, then we can pick $\delta=\frac{\epsilon}{4}$ which gives us $\delta^2+2\delta<\epsilon$. If $\epsilon \ge8$, then we can pick $\delta=2$ which gives us $\delta^2+2\delta\le\epsilon$.

Therefore, if we pick $\delta =$ min {${\frac{\epsilon}{4}, 2}$}, then $0 < \sqrt {(x-1)^2+(y-1)^2}<\delta$ implies $0<|xy-1|<\epsilon$

Thus, $\lim\limits_{(x,y) \to (1,1)} xy=1$

Answer 1

A shorthand you might find useful:
Since $$(x-y)^2 \gt 0 \Rightarrow xy \lt \frac{x^2+y^2}{2} $$ Also, note that $$\lVert (x-1, y-1) \rVert \lt \delta \Rightarrow |x-1| \lt \delta \;\land\; |y-1| \lt \delta \Rightarrow x \lt \delta+1 \;\land\; y \lt \delta+1 $$

Answer 2

Your proof is correct and nice.

Take $\delta<1$

By doing this a priori you will find the delta you need more easily and save some time from the calculations

You proved that $$|xy-1| \leq |x-1||y-1|+|x-1|+|y-1|$$

Thus $$|x-1|=\sqrt{(x-1)^2} \leq \sqrt{(x-1)^2+(y-1)^2} < \delta$$ $$|y-1|=\sqrt{(y-1)^2} \leq \sqrt{(x-1)^2+(y-1)^2} < \delta$$

Thus $|f(x,y)-1|=|xy-1| <\delta^2 +2\delta=\delta(\delta+2) <3 \delta$

Choose $0<\delta <\min\{ \frac{\epsilon}{3},1\}$ and you have the result