I am currently working on limits. In an early edition of "Calculus", Spivak uses the following lemma to proof the multiplication law of limits:
Lemma: If $|x - x_0| < \min(1, \frac{\epsilon}{2|y_0|+1})$ and $|y - y_0| < \frac{\epsilon}{2|x_0|+1})$, then $|xy - x_0y_0| < \epsilon$.
Proof: Since $|x - x_0| < 1$ we have
$$|x| - |x_0| \le |x - x_0| < 1,$$
so that
$$|x| < 1 + |x_0|.$$
Thus
$$\begin{align}|xy - x_0y_0| & = |x(y-y_0) + y_0(x-x_0)| \\& \le |x| \cdot |y-y_0| + |y_0| \cdot |x-x_0| \\& < (1 + |x_0|) \cdot \frac{\epsilon}{2|x_0|+1} + |y_0| \cdot \frac{\epsilon}{2|y_0|+1} \\& = \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon\end{align}$$
$\blacksquare$
Spivak uses this to proof that, if $\lim_{x \to a}f(x)=l$ and $\lim_{x \to a}g(x)=m$ then $\lim_{x \to a}(f \cdot g)(x) = l \cdot m$ by corresponding $f(x)$ with $x$, $l$ with $x_0$, $g(x)$ with $y$ and $m$ with $y_0$.
My question is the following:
Is it necessary, to require that $|x - x_0| < \min(1, \frac{\epsilon}{2|y_0|+1})$? I think it would suffice if $|x - x_0| < \min(1, \frac{\epsilon}{2|y_0|})$. Even in applying the lemma to the actual proof, since $\lim_{x \to a}f(x)=l$ exists it should give no more problems to find a $\delta$ for $\frac{\epsilon}{2|y_0|}$ than for $\frac{\epsilon}{2|y_0|+1}$.
Is this a typo in the textbook or am I missing something?