Help with Spivak's Calculus: Chapter 1 problem 21 I've been stuck on this problem for over a day, and the answerbook simply says "see chapter 5" for problems 20,21, and 22. But I want to complete the problem without using knowledge given later in the book, so I've been banging my head against the wall trying all sorts of things, but nothing I do seems to lead me anywhere.
The problem is as follows:
Prove that if
$|x - x_0| < min (\frac{\varepsilon}{2(|y_0| + 1)}, 1)$ and $|y - y_0| < min (\frac{\varepsilon}{2(|x_0| + 1)}, 1)$
then
$|xy - x_0 y_0| < \varepsilon$.
Here are some of the things I've been thinking about, I don't know which of these are useful (if any), but they somewhat outline the logic behind my various attempts.
Since at most $|x - x_0| < 1$ and $|y - y_0| < 1$ then it follows that $(|x-x_0|)(y-y_0|) < |x-x_0|$ and $(|x-x_0|)(y-y_0|) < |y-y_0|$
Also, $(|x - x_0|)(|y_0| + 1) < \frac{\varepsilon}{2}$ and $(|y - y_0|)(|x_0| + 1) < \frac{\varepsilon}{2}$ so $(|x - x_0|)(|y_0| + 1) + (|y - y_0|)(|x_0| + 1) < \varepsilon$. and since $|a + b| \leq |a| + |b| < \varepsilon$ I've tried multiplying things out, and then adding them together to see if anything cancels, but I can't make anything meaningful come out of it.
Also since $|a - b| \leq |a| + |b|$ I've also tried subtracting one side from the other, but to no avail.
I was also thinking that since $(|x-x_0|)(y-y_0|) < |x-x_0|$, then I could try something along the lines of $(|x-x_0|)(y-y_0|)(|x_0| + 1) + (|x-x_0|)(y-y_0|)(|y_0| + 1)< \varepsilon$ and various combinations as such, but I just can't seem to get anything meaningful to come out of any of these attempts.
I have a sneaking suspicion that the road to the solution is simpler than I'm making it out to be, but I just can't see it. 
 A: By reverse triangle inequality $|x| - |x_0| \leq |x-x_0| < 1$ giving $|x| < 1 + |x_0|$. Now 
\begin{align*}
|xy - x_0y_0| & = |x(y-y_0) + y_0(x-x_0)| \leq |x||y-y_0| + |y_0||x-x_0| \\ & < (1 + |x_0|)\frac{\varepsilon}{2(|x_0| + 1)} + |y_0|\frac{\varepsilon}{2(|y_0| + 1)} \\ & < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon
\end{align*}
To my mind the difficult part of this argument is writing down the equality proceeding the string of inequalities and this certainly isn't the only way to do that.
A: I just solved this today - which for me was really amazing because I was just as lost as the earlier post - so please forgive my enthusiasm. The breakthrough for me was the hint $|x - x_0|$ < 1 and a relaxation of mind. 
$|y - y_0|$ < ${\frac{\epsilon}{2(|x_0| + 1)}}$ < ${\frac{\epsilon}{2(|x_0| + |x - x_0|)}}$.
This was the application of the hint.
$|y - y_0|$ < ${\frac{\epsilon}{2(|x_0| + |x - x_0|)}}$ < ${\frac{\epsilon}{2(|x_0 + x - x_0|)}}$
$|y - y_0|$ < ${\frac{\epsilon}{2|x|}}$
(1) $|x||y - y_0|$ < ${\frac{\epsilon}{2}}$ 
Working with the given inequality:
$|x - x_0|$ < ${\frac{\epsilon}{2(|y_0| + 1)}}$
Simplifying:
(2) $|x - x_0|(|y_0| + 1) < {\frac{\epsilon}{2}}$
Add (1) and (2) following: if a < b and c < d, then a+c < b+d.
$|x||y - y_0| + |x - x_0|(|y_0| + 1) < \epsilon$
leads to:
$|xy - xy_0| + |x-x_0||y_0| + |x - x_0| < \epsilon$
$|xy - xy_0 + xy_0 - x_0y_0| + |x - x_0|$ < $|x||y - y_0| + |x - x_0|(|y_0| + 1) < \epsilon$
$|xy - x_0y_0| + |x - x_0| < \epsilon$
$|xy - x_0y_0| < |xy - x_0y_0| + |x - x_0| < \epsilon$
$|xy - x_0y_0| < \epsilon$
A: I think I figured it out thanks to GeoffRobinson's hint, and from RJS (thanks guys!) so I figured I'd write out my own work here.
$|xy-x_0y_0| = |x(y-y_0) + y_0(x-x_0)| \leq |x(y-y_0)| + |y_0(x-x_0)|$
So in essence we're going to show that $|x(y-y_0)| + |y_0(x-x_0)| \leq \varepsilon$ which implies that $|xy-x_0y_0| \leq \varepsilon$.
So let's start. From
$|x - x_0| < \frac{\varepsilon}{2(|y_0| + 1)}$
then
$(|x - x_0|)(|y_0| + 1) = |y_0(x-x_0)| + |x - x_0| < \frac{\varepsilon}{2}$
Which in turn implies that $|y_0(x-x_0)|< \frac{\varepsilon}{2}$
We're halfway done with our inequality.
Next by looking at 
$|x-x_0| < 1$, we can observe that $|x| - |x_0| \leq |x-x_0| \Rightarrow |x| < |x_0| + 1$
Then consider the given inequality:
$|y-y_0| < \frac{\varepsilon}{2(|x_0| + 1)}$
We can see that $|x||y-y_0| < (|x_0| + 1) \frac{\varepsilon}{2(|x_0| + 1)} = \frac{\varepsilon}{2}$ and so $|x(y-y_0)| < \frac{\varepsilon}{2}$
So finally we can show that:
$|y_0(x-x_0)| + |x(y-y_0)| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon$
and because $|xy-x_0y_0| \leq |x(y-y_0)| + |y_0(x-x_0)|$
We can say $|xy-x_0y_0| < \varepsilon$
