Spivak Chapter 5, Lemma 3 - Explanation I'm having some trouble understanding the 3rd lemma in Chapter 5 (page 101 for the 3rd edition) of Spivak's Calculus, that states:
If 
$\ y_{0} \neq 0 $ and $\ |y-y_{0}|  < min(\frac{|y_{0}|}{2},  \frac{\epsilon|y_{0}|^2}{2}) $
then
$\ y \neq 0 $  and $\ \left |\frac{1}{y}-\frac{1}{y_{0}} \right | < \epsilon $
So naturally I am not understanding the proof either.
I don't understand why he uses $\ |y-y_{0}| $ instead of $\ |x-x_{0}| $ --  EDIT I think I figured what he means with that, however I am not sure as to why y instead of x; regardless, am I correct in thinking of $\ y $ as a separate function such as $\ f(x) $ and $\ y_{0} $ as the limit of said function? -- and whether I am right in assuming for the proof that the reason he uses $\ \frac{|y_{0}|}{2} $ as the $ \min $ is because $\ y_{0}^2 $ >> $\ y_{0} $ regardless of the value of $\ \epsilon $.
Can anyone help me out?
 A: Answering so as to have a nice excuse to flip through Spivak's book :).
To your first question, no, you are not correct in thinking that $y=f(x)$. While I guess this notation is common e.g. in differential equation material you might have seen, here $y_0$ is an arbitrary number $\neq 0$ and so is $y$. Of course he could have named these $x_0$, $x$ or $z_0, z$ or in any other imaginable way. So if it makes it easier for you just imagine $x_0$ and $x$ instead of $y_0$ and $y$. The fact that $y$ is not a function of $x$ should become clearer from the other statement in the same Lemma, where he shows that $|xy-x_0y_0| < \varepsilon$, for $x,y$ in appropriate neighborhoods of $x_0$ and $y_0$ respectively.
To your second question, here you are also wrong. In fact you can easily check that $y_0^2 > y_0$ is a wrong statement (and $\varepsilon$ does not even play any role here). I really think you should recheck the definition of the limit to understand what is happening. Here Spivak basically shows that $ \lim_{y\to y_0} \frac{1}{y} = \frac{1}{y_0}$, assuming $y_0 \neq 0$. This means that given $\varepsilon > 0$ he has to find $\delta > 0$ such that $\big\vert\frac{1}{y} - \frac{1}{y_0}\big\vert < \varepsilon$ for all $y$ with $|y-y_0| < \delta$. For fixed $\varepsilon > 0$ he then goes on to choose:
$$\delta = \min\left(\frac{|y_{0}|}{2},  \frac{\varepsilon|y_{0}|^2}{2}\right)$$
Notice first that $\delta > 0$ (why?) and that by the definition of $\delta$ you have that: 
1) $\delta \leq \frac{|y_{0}|}{2}$
2) $\delta \leq \frac{\varepsilon|y_{0}|^2}{2}$
The first equation is needed in the first part of his proof, while the second in the final step of the proof, which essentially shows that for this choice of $\delta>0$ it holds that $\big\vert\frac{1}{y} - \frac{1}{y_0}\big\vert < \varepsilon$ for all $y$ in the $\delta$-neighborhood of $y_0$.
Hope this helps.
A: Lemma (3) is stated as:
If $ y_0 \neq 0$ and 
$$ |y - y_0| < \text{min}(\frac{|y_0|}{2}, \frac{\varepsilon|y_0|^2}{2})
$$
then $y \neq 0$ and
$$ |\frac{1}{y} - \frac{1}{y_0}| < \varepsilon $$
Here $y$ and $y_0$ act the same as $x$ and $x_0$: that is, $y$ and $y_0$ are numbers and we are interested in the distance between them and how close they are. Spivak's choice of variable here is, as far as I can tell, completely arbitrary.
Now for the proof:
We have
$$ |y| -|y_0| \leq |y - y_0| < \frac{|y_0|}{2} $$
This part is stating that if you take the distance of any two numbers from zero and subtract them, that difference will be less than the distance from 0 of the difference of the numbers. That fact gives us the first half of the inequality. The second portion $|y - y_0| < \frac{y_0}{2}$ comes from the given that $|y - y_0| < \text{min}(\frac{|y_0|}{2}, \frac{\varepsilon|y_0|^2}{2})$. So we have
$$|y_0| - |y| < \frac{|y_0|}{2} \\
|y_0| - \frac{|y_0|}{2}  < |y| \\
|y| > \frac{|y_0|}{2}
$$
The smallest value that $\frac{|y_0|}{2}$ can take is $0$, so we have $|y| > 0$, and thus more importantly $ y \neq 0$. From this same inequality and some algebraic manipulation we have 
$$ \frac{1}{|y|} < \frac{2}{|y_0|}
$$
Now consider
$$|\frac{1}{y} - \frac{1}{y_0}| = \frac{|y - y_0|}{|y|} \cdot |y_0| = \frac{1}{y}\cdot \frac{1}{y_0} \cdot |y - y_0|$$
From the given inequalities and the ones we have shown, we have 
$$\frac{1}{y}\cdot \frac{1}{y_0} \cdot \frac{\varepsilon|y_0|^2}{2} < \frac{2}{y_0}\cdot \frac{1}{y_0} \cdot \frac{\varepsilon|y_0|^2}{2} = \varepsilon$$
So $|\frac{1}{y} - \frac{1}{y_0}| < \varepsilon$ $\blacksquare$
The cancellation in the last line that leaves us with $\varepsilon$ may seem too good to be true and indeed it is: the given condition of $ |y - y_0| < \text{min}(\frac{|y_0|}{2}, \frac{\varepsilon|y_0|^2}{2})$ guarantees it will cancel nicely. That is why some of those terms may seem like odd choices at the beginning. In reality, if you were proving something like this, you would likely fiddle around in the middle a bit and then go back to the start after realizing what terms are needed to make things cancel in the end.
