# I not understanding a cancellation step in Spivak proof.

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.

• Note that the content of next line would imply that $\frac{\varepsilon}{2}\frac{|y_0|}{|y_0|+1}<\frac{\varepsilon}{2}$. This is true because $|y_0|<|y_0|+1$ and consequently $\frac{|y_0|}{|y_0|+1}<1$. Commented Jun 12, 2022 at 2:31
• Thanks for that explanation @AlannRosas . That was helpful. Commented Jun 12, 2022 at 2:47

$$\vert y_0 \vert \cdot \frac{\epsilon}{2(\vert y_0 \vert + 1 )} = \frac{\epsilon}{2}$$
$$\vert y_0 \vert \cdot \frac{\epsilon}{2(\vert y_0 \vert + 1 )} < \frac{\epsilon}{2}$$
which follows from $$|y_0|<|y_0|+1$$.
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.