Spivak's Calculus - Chapter 1 Question 23 In Spivak's Calculus Chapter 1 Question 23:

Replace the question marks in the following statement ing $\varepsilon, x_0$ and $y_0$ so that the conclusion will be true:
  If $y_0\neq 0$ and $$|y-y_0|<?  \qquad\text{and}\qquad  |x-x_0|<?$$ then $y\neq 0$ and $$ \bigg| \frac{x}{y}-\frac{x_0}{y_0}\bigg|<\varepsilon.$$

The answer in its Solution Manual is $$|x-x_0|<\min\bigg(\frac{\varepsilon}{2(1/|y_0|+1)},1 \bigg)$$ and $$|y-y_0|<\min\bigg( \frac{|y_0|}{2},\frac{\varepsilon|y_0|^2}{4(|x_0|+1)} \bigg).$$ The latter is same as my solution, but the first is a bit different. So I wonder if $$|x-x_0|<\min\bigg(\frac{\varepsilon |y_0|}{4},1 \bigg)$$, is the solution still be true?
Thanks in advance.
 A: $|x-x_0|<\min\bigg(\frac{\varepsilon |y_0|}{4},1 \bigg)$ is wrong, here's why:
We have:
\begin{align*}
\bigg| \frac{x}{y}-\frac{x_0}{y_0}\bigg| &\leq \frac{\left | x \right |\left | y_{0}-y \right |+\left | y \right |\left | x_{0}-x \right |}{\left | yy_{0} \right |}\\
&\leq \frac{\left | x-x_{0} \right |\left | y-y_{0} \right |}{\left | yy_{0} \right |}+\frac{\left | x_{0} \right |\left | y_{0}-y \right |}{\left | yy_{0} \right |}+\frac{\left | y \right |\left | x_{0}-x \right |}{\left | yy_{0} \right |}
\end{align*}
where I used: $\left | x \right |\leq \left | x-x_{0} \right |+\left | x_{0} \right |$
Now if you consider the first term, i.e. $\frac{\left | x-x_{0} \right |\left | y-y_{0} \right |}{\left | yy_{0} \right |}$ alone and try to get it to be smaller than $\frac{\varepsilon }{2}$ you would end up with:
$\frac{\left | x-x_{0} \right |\left | y-y_{0} \right |}{\left | yy_{0} \right |}\leq \frac{\left | x-x_{0} \right |\frac{y_{0}}{2}}{\frac{\left | y_{0} \right |^{2}}{2}}= \frac{2\left | x-x_{0} \right |}{\left | y_{0} \right |}< \frac{\varepsilon }{2}$ giving the result you found which is $|x-x_0|<\frac{\varepsilon |y_0|}{4}$.
But looking at the decomposition above you realise this is wrong because you need to consider: 
$\frac{\left | x-x_{0} \right |}{\left | yy_{0} \right |}(\left | y-y_{0} \right |+\left | y \right |)$ instead of $\frac{\left | x-x_{0} \right |\left | y-y_{0} \right |}{\left | yy_{0} \right |}$ 
The correct answer for the first is the one stated by the manual as:
$\left | \frac{x}{y}-\frac{x_{0}}{y_{0}} \right |=\left | \frac{x}{y}-\frac{x}{y_{0}}+\frac{x}{y_{0}}-\frac{x_{0}}{y_{0}} \right |\leq \left | x \right |\left | \frac{1}{y}-\frac{1}{y_{0}} \right |+\left |\frac{1}{y_{0}}  \right |\left | x-x_{0} \right |$
So:
$\left |\frac{1}{y_{0}}  \right |\left | x-x_{0} \right |\leq \frac{\frac{1}{\left |y_{0}  \right |}\varepsilon }{2(\frac{1}{\left |y_{0}  \right |}+1)}\leq \frac{\varepsilon }{2}$
