What do the denominator mean in this theorem in Spivak's book

Question

In Spivak's Calculus there is problem 21 in chapter 1. which says:

Prove that if

$|x - x_0| <\min \Bigl(\frac{\varepsilon}{2(|y_0| + 1)}, 1\Bigr)$ and $|y - y_0| < \frac{\varepsilon}{2(|x_0| + 1)}$

then

$|xy - x_0 y_0| < \varepsilon$.

I'd like to at least follow along some of the proofs I found for this, because I know it's used later in the book. But I get confused, because I can't really understand what's the significance is of $2(|x_0| + 1)$ and $2(|y_0| + 1)$.

I know the theorem used for delta/epsilon later on, but I'd like to get an intuitive understanding.

I found one intuition behind the theorem:

If you have a rectangle with sides lengths $x_0$ and $y_0$ and you want to lengthen or shorten each side in such way that area of the rectangle changes by no more than $\epsilon$, then you can do so by changing $x_0$ side by less than $\text{min}(1, \frac{\epsilon}{2(y_0 + 1)})$ and $y_0$ by less than $\frac{\epsilon}{2(x_0 + 1)}$.

But unfortunately I can't see it. How does it relate to a rectangle?

Maybe someone has another explanation of what we are trying to proof in this theorem.

Answer 1

This is just a technical lemma on the way to prove that the limit of the product is the product of the limits.

If you want to add a little bit more of geometric intuition to it, consider the following case. Assume that there is a rectangle with one of the corners in origin and sides of lengths $x_0$ and $y_0$. Ask the following question -- how can you increase lengths of sides $x_0$ and $y_0$ in order to increase the total area no more than by $\varepsilon$?

The area of the rectangle is given by $S = x_0\cdot y_0$, so the increment is $\Delta S = \Delta x_0\cdot \Delta y_0$. Now lets check that $\Delta x \cdot \Delta y$ is indeed less or equal than $\varepsilon$.

$$\Delta x \cdot \Delta y = \min(1, \frac{\varepsilon}{2(y_0+1)})\cdot\frac{\varepsilon}{2(x_0+1)}=\min(\frac{\varepsilon}{2(x_0+1)},\frac{\varepsilon^2}{4(y_0+1)(x_0+1)})$$

Is it less than $\varepsilon$? Sure! In first case, $\frac{\varepsilon}{2(x_0+1)}$ is less than $\varepsilon$ because denominator is a positive number greater than $1$, and in the second case, $\frac{\varepsilon^2}{4(y_0+1)(x_0+1)}$ is less than $\varepsilon$ because it is a product of $\frac{\varepsilon}{2(x_0+1)}$ with $\frac{\varepsilon}{2(y_0+1)}$, which is less than $1$.

Answer 2

$|xy-x_0y_0| < \epsilon$

Adding and subtracting the same term doesn't change the expression, but it will allow us to factor.

$|xy-x_0y + x_0y - x_0y_0|< \epsilon\\ |(x-x_0)y - x_0(y-y_0)|< \epsilon$

By the triangle inequality

$|(x-x_0)y - x_0(y-y_0)| \le |(x-x_0)y| + |x_0(y-y_0)|$

if $|(x-x_0)y|<\frac {\epsilon}{2}$ and $|x_0(y-y_0)| < \frac {\epsilon}{2}$ then $|xy-x_0y_0| < \epsilon$

and that will tell us how close $(x,y)$ must be to $(x_0,y_0)$

$|x-x_0| < \frac {\epsilon}{2|y|}$ and $|y-y_0| < \frac {\epsilon}{2|x_0|} $

We don't want that $y$ in the denominator. We want an expression in terms of $y_0$ We also need to cover our bases in the case that either $x_0$ or $y_0 = 0$

Lets assume that the distance between $(x,y)$ and $(x_0,y_0)$ is less than 1

$|x - x_0| < 1$ and $|y|< |y_0|+1$ and in all cases is $|x_0| < |x_0| + 1$

and

$|x-x_0| < \min(1,\frac {\epsilon}{2(|y_0| + 1)})$ and $|y-y_0| \le \frac {\epsilon}{2(|x_0| + 1)}$ allow us to say that there is a delta such that $d((x,y),(x_0,y_0))<\delta \implies|xy-x_0y_0| < \epsilon$