I not understanding a cancellation step in Spivak proof.

Question

There is a question about this exact same proof but the answer is not satisfying me so I'm going to run it again if you don't mind. I will post the original question after mine. In his chapter on limits, Spivak uses the following lemma:

$$\left\lvert x - x_0 \right\rvert < \min\left(1, \frac{\epsilon}{2(\lvert y_0 \rvert+1)}\right)$$ and $$\lvert y - y_0 \rvert \lt \frac{\epsilon}{2(\lvert x_0\rvert + 1) }$$ then $$\lvert xy - x_0y_0\rvert < \epsilon$$

So here is his proof:

Since we have $\lvert x - x_0 \rvert < 1$ we have $$\lvert x \rvert - \lvert x_0 \rvert \leq \lvert x - x_0 \rvert \lt 1$$ so that $$\vert x \vert \lt 1 + \vert x_0 \vert$$ Thus $$ \begin{aligned} \vert xy - x_0y_0\vert &= \vert x(y-y_0) + y_0(x - x_0) \vert \\ &\le \vert x \vert \cdot \vert y - y_0 \vert + \vert y_0 \vert \cdot \vert x - x_0 \vert\\ &\lt (1 + \vert x_0 \vert) \cdot \frac{\epsilon}{2(\vert x_0 \vert + 1)} + \mathbf{ \vert y_0 \vert \cdot \frac{\epsilon}{2(\vert y_0 \vert + 1 )}} \\ &\lt \frac{\epsilon}{2} + \mathbf{\frac{\epsilon}{2}} = \epsilon \end{aligned} $$

I have bolded the transformation from the second-last line to the last line that I do not understand.

How do we get

$$ \vert y_0 \vert \cdot \frac{\epsilon}{2(\vert y_0 \vert + 1 )} = \frac{\epsilon}{2}$$

Not understanding a cancellation step in an inequality proof from Spivak's Calculus.

Answer 1

How do we get

$$ \vert y_0 \vert \cdot \frac{\epsilon}{2(\vert y_0 \vert + 1 )} = \frac{\epsilon}{2}$$

This isn't what he used there, he used

$$ \vert y_0 \vert \cdot \frac{\epsilon}{2(\vert y_0 \vert + 1 )} < \frac{\epsilon}{2}$$

which follows from $|y_0|<|y_0|+1$.

Answer 2

What's being used is $$ \frac{|y_0|}{|y_0|+1}< 1 $$since multiplying both sides by $\frac{\varepsilon}{2}>0$ gives the last step. The inequality is true since $$ \frac{|y_0|}{|y_0|+1}< 1 \iff |y_0| < |y_0|+1 \iff 0 < 1 $$ which is true.