Spivak's Calculus (4th ed): Chapter 1 Problem *21 Inequality

[Steve Turchin]

Homework Statement

Prove that if
$|x-x_0|<\min (\frac {\epsilon}{2(|y_0|+1)},1)$ and $|y-y_0|<\frac{\epsilon}{2(|x_0|+1)}$

then
$|xy-x_0y_0|<\epsilon$

Homework Equations

N/A

The Attempt at a Solution

From the first inequality I can see that:
$|x-x_0|<\frac {\epsilon}{2(|y_0|+1)}$ and $|x-x_0|<1$
From the first and second inequalities:
$|x-x_0|(2(|y_0|+1))<\epsilon$
$|y-y_0|(2(|x_0|+1))<\epsilon$
so by adding up both inequalities I get:
$|x-x_0|+|y_0|\cdot|x-x_0|+|y-y_0|+|x_0|\cdot|y-y_0|<\epsilon$
and now I know that I need to write $xy-x_0y_0$ in a way that involves $x-x_0$ and $y-y_0$

Thanks in advance!

 

[Samy_A]

$|x-x_0|+|y_0|\cdot|x-x_0|+|y-y_0|+|x_0|\cdot|y-y_0|<\epsilon$

Building on this, a hint:
$xy-x_0y_0=xy-xy_0+xy_0-x_0y_o=x(y-y_0)+y_0(x-x_0)$

Now the $|x||y-y_0|$ term doesn't appear in your expression, but notice that $x=x-x_0+x_o$, and remember that $|x-x_0|<1$