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.