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

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


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


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



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

  • $\begingroup$ Personal laziness is keeping me from going through all the details here, but even something like this is a good example of why developing some basic theory is so useful. It's easy enough to show $f(x,y) = x$ and $g(x,y) = y$ are continuous from the definition. Then you prove that the product of continuous functions is continuous, which gives you the above limit immediately. $\endgroup$ – Christian Sykes Sep 21 '17 at 6:05
  • $\begingroup$ I'm not saying there's no value in playing with $\epsilon$s and $\delta$s, but it can definitely be a case of diminishing returns. Things get unruly pretty quickly. $\endgroup$ – Christian Sykes Sep 21 '17 at 6:07
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    $\begingroup$ In $\mathbb{R}^n$ all norms are equivalent so i recomment to use the $\|.\|_\infty$ this will make your calculation easier. $\endgroup$ – Nathanael Skrepek Sep 21 '17 at 6:08
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    $\begingroup$ Very very good proof, honestly. I am very happy to see that you have passed the rigour test with flying colours. You will be delighted to know that in future you will develop mathematical tools that will make this problem a mere formality. One of these is suggested in the comment above,namely that you can change norm from the cumbersome Euclidean one to the infinity norm. $\endgroup$ – астон вілла олоф мэллбэрг Sep 21 '17 at 6:12

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


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


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