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Preliminaries:

$$|x - x_0| < \min \left(1, \frac{\varepsilon}{2(|y_0| + 1)}\right)$$

$$|y - y_0| < \frac{\varepsilon}{2(|x_0| + 1)}$$

Proof:

$$|xy - x_0y_0| = |x(y-y_0) + y_0(x-x_0)|$$ $$\qquad \qquad \qquad \ \ \le |x| \cdot |y-y_0| + |y_0| \cdot |x-x_0|$$ $$\qquad \qquad \qquad \qquad \qquad \qquad \ \ \lt (1 + |x_0|) \cdot \frac{\varepsilon}{2(|x_0| + 1)} + |y_0| \cdot \frac{\varepsilon}{2(|y_0| + 1)}$$ $$< \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon $$

My only hurdle in this proof understanding is how Spivak goes from

$|x| \cdot |y-y_0| + |y_0| \cdot |x-x_0|$

to

$|x| \cdot |y-y_0| + |y_0| \cdot |x-x_0| \lt (1 + |x_0|) \cdot \frac{\varepsilon}{2(|x_0| + 1)} + |y_0| \cdot \frac{\varepsilon}{2(|y_0| + 1)}$

It's clear that we are using the preliminaries here. But where did the $\min$ go. Just full working out of steps here would be very appreciated.

Sebastiano
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Naz
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  • $|x-x_{0}|\leq 1$ so $|x| - |x_{0}| \leq 1$. The min is just a short way of writing $|x-x_{0}|$ is bounded by both of those quantities. – FZan Jan 26 '24 at 03:31
  • looked at: https://math.stackexchange.com/questions/225609/prove-that-xy-x-0y-0-epsilon?rq=1 – Naz Jan 26 '24 at 11:42
  • and they use the fact that you mentioned @FZan, that $|x| \leq (1 + |x_0|)$. This whole equation makes sense now – Naz Jan 26 '24 at 11:44

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