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.
Best Answer
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 $|y_0|<|y_0|+1$.