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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.

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    $\begingroup$ Hint: $(xy-x_{0}y_{0}) = x(y-y_{0}) + y_{0}(x-x_{0})$. $\endgroup$ – Geoff Robinson Aug 6 '15 at 20:25
  • $\begingroup$ @GeoffRobinson Thanks for the hint. I'll try to see if I can put it to use. $\endgroup$ – Shane T Aug 6 '15 at 20:30
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    $\begingroup$ The game is given away in the answer below, although if you follow the hint, there is a natural progression to the necessary inequalities. $\endgroup$ – Geoff Robinson Aug 6 '15 at 21:42
  • $\begingroup$ @GeoffRobinson my only remaining question is how or why you recognized that (xy−x_0y_0)=x(y−y_0)+y_0(x−x_0)? I don't know if I would have seen that without you telling me. $\endgroup$ – Shane T Aug 6 '15 at 23:17
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    $\begingroup$ Some things you just have to see- they come with experience- the intuition is that you know that if $x$ is close to $x_{0}$ and $y$ is close to $y_{0}$, then $xy$ is close to $x_{0}y_{0}$. But to make that a reality, you have to see that step. If you want to break it down further, you know that $x(y-y_{0})$ is small, but that isn't $xy - x_{0}y_{0}$. To get that, you have to add $xy_{0}- x_{0}y_{0} = y_{0}(x-x_{0}).$ $\endgroup$ – Geoff Robinson Aug 6 '15 at 23:47
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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.

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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:

$|xy - xy_0| + |x-x_0||y_0| + |x - x_0| < \epsilon$

$|xy - xy_0 + xy_0 - x_0y_0| + |x - x_0|$ < $|x||y - y_0| + |x - x_0|(|y_0| + 1) < \epsilon$

$|xy - x_0y_0| + |x - x_0| < \epsilon$

$|xy - x_0y_0| < |xy - x_0y_0| + |x - x_0| < \epsilon$

$|xy - x_0y_0| < \epsilon$

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  • $\begingroup$ Nice! I'm glad you were able to figure it out. It took me a while too. $\endgroup$ – Shane T Aug 17 '15 at 18:35
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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 $|xy-x_0y_0| < \varepsilon$

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