# Difficulty understanding particular proof on limit properties (not epsilon/delta)

1. Jan 14, 2010

### alyks

Hi, I've been studying from Spivak's Calculus. Normally when I have trouble I can just search, but this time I can't find anything (you can tell how extensive this forum is in that I've been registered here for a while and this is my first post). On page 89, the book gives proof of the following:

If
$$|x - x_0| < \text{min} \left (1, \frac{\epsilon}{2(|y_0| + 1)}\right)$$ and $$|y - y_0| < \frac{\epsilon}{2(|x_0| + 1)}$$

Then

$$|xy - x_0y_0| < \epsilon$$

The proof shows this:

1. Since $$|x-x_0| < 1$$ we have $$|x| - |x_0| \leq |x-x_0| < 1$$ so that $$|x| < 1 + |x_0|$$

Thus,

2. $$|xy - x_0y_0| = |x(y-y_0) + y_0(x-x_0)|$$
3. $$\leq |x| \cdot |y - y_0| + |y_0| \cdot (|x - x_0|)$$
4. $$\leq (1 + |x_0|) \cdot \frac{\epsilon}{2(|x_0| + 1)} + |y_0| \cdot \frac{\epsilon}{2(|y_0| + 1)} = \frac{\epsilon}{2} + \frac{\epsilon}{2}$$

Where I have a problem is in how he just assumes that $$\text{min} \left (1, \frac{\epsilon}{2(|y_0| + 1)}\right)$$ is 1, when I saw the minimum I would have thought you'd do a proof by cases.

Then later, I have a hard time understanding how he went from line 3 to line 4. If $$|x| < 1 + |x_0|$$ and $$|y - y_0| < \frac{\epsilon}{2(|x_0| + 1)}$$, then only half of line four makes sense. Otherwise, I'm lost. Will anybody help me out?

2. Jan 14, 2010

### snipez90

He doesn't assume that $\text{min} \left (1, \frac{\epsilon}{2(|y_0| + 1)}\right)$ is 1. If epsilon is small enough, then clearly the minimum will not be 1. In any case, you'll save yourself a lot of confusion if you think of |x - a| < min(y,z) as simply stating that both |x-a| < y AND |x-a| < z are satisfied. In fact, look back at exercises 22-24 (or somewhere around there) of chapter 1; he states this reasoning explicitly. I'm not sure what's confusing about going from line 3 to 4... If you are wondering why there is an equals sign in line 4, I'm pretty sure that is a minor oversight. It should be a less than sign since |y_0| < |y_0| + 1.

3. Jan 14, 2010

### alyks

Oh wow, it does save a lot of confusion with the min, I was thinking about it wrong.

4. Jan 16, 2010

### evagelos

1st of all No 4 is wrong,because :

$$(1 + |x_0|) \cdot \frac{\epsilon}{2(|x_0| + 1)} + |y_0| \cdot \frac{\epsilon}{2(|y_0| + 1)}$$

IT is not equal to ε/2 + ε/2,

but :

$$(1 + |x_0|) \cdot \frac{\epsilon}{2(|x_0| + 1)} + (|y_0|+1) \cdot \frac{\epsilon}{2(|y_0| + 1)}$$ it is equal to ε/2 +ε/2

So you have to extend No 3 inequality and :

3. $$\leq |x| \cdot |y - y_0| + |y_0| \cdot (|x - x_0|)$$ $$\leq |x| \cdot |y - y_0| + (|y_0|+1) \cdot (|x - x_0|)$$.

Then :

$$|x - x_0| < \text{min} \left (1, \frac{\epsilon}{2(|y_0| + 1)}\right)$$ and $$|y - y_0| < \frac{\epsilon}{2(|x_0| + 1)}$$

Implies that:

1) $$|x-x_{o}|\leq 1$$ which implies that: $$|x|\leq |x_{o}| +1$$

AND

2) $$|x-x_{o}|\leq\frac{\epsilon}{2(|y_{o}|+1)}$$
AND

3) $$|y-y_{o}|<\frac{\epsilon}{2(|x_{o}|+1)}$$

AND using (1) (2) and (3) we have :

$$|xy-x_{o}y_{o}|\leq |x| \cdot |y - y_0| + (|y_0|+1) \cdot (|x - x_0|)<(1 + |x_0|) \cdot \frac{\epsilon}{2(|x_0| + 1)} + (|y_0|+1) \cdot \frac{\epsilon}{2(|y_0| + 1)}= \frac{\epsilon}{2}+\frac{\epsilon}{2} = \epsilon$$