have just been reading through and doing the questions in Spivak's Calculus, but am not entirely sure I am understanding a step in one of the proofs given in chapter $5$ (but was given as an exercise in the first chapter).
I had found a picture of the question online, but apparently I can't post it here. The proof is on page $89$, chapter $5$, and its for the $2nd$ lemma.
Lemma: If $|x-x_0|<\min\left(1,\frac{\varepsilon}{2(|y_0|+1)}\right)$ and $|y-y_0|<\frac\varepsilon{2(|x_0|+1)}$, then $|xy-x_0y_0|<\varepsilon$
Proof: $$|xy-x_0y_0|=|x(y-y_0)+y_0(x-x_0)|\le|x||y-y_0|+|y_0||x-x_0|$$ then \begin{align}&|x||y-y_0|+|y_0||x-x_0|<(1+|x_0|)\frac\varepsilon{2(|x_0|+1)}+|y_0|\frac{\varepsilon}{2(|y_0|+1)}\\ &=\frac\varepsilon2+\frac\varepsilon2=\varepsilon\end{align}
In the second to last line of the proof, I don't understand how the $\dfrac{\varepsilon \lvert y_0\rvert}{2(\lvert y_0\rvert+1)}$ is meant to cancel to $\dfrac\varepsilon2$.
Did they use $\dfrac{\lvert y_0\rvert}{\lvert y_0\rvert+1}<1$ so $\dfrac{\varepsilon \lvert y_0\rvert}{2(\lvert y_0\rvert+1)}<\dfrac\varepsilon2$? But if they did something like this, shouldn't that last line then have a "$<$" instead of a "$=$"?
