The following problem is from Spivak's Calculus (4th ed., pg. 18):

Prove that if:

$|x-x_0| < \min(\frac{\epsilon}{2(|y_0|+1)}, 1)$ and $|y - y_0| < \frac{\epsilon}{2(|x_0|+1)}$ then $|xy-x_0 y_0| <\epsilon$.

What I've done so far is to say that the since the first inequality implies $|x-x_0| < 1$, combining this with the second inequality yields $|(x-x_0)(y-y_0)|<\frac{\epsilon}{2(|x_0|+1)}$, or $|(x-x_0)(y-y_0)|\cdot(|x_0|+1) < \frac{\epsilon}{2}$. We also know from the first inequality in the problem that $|x-x_0|<\frac{\epsilon}{2(|y_0|+1)}$, so $|x-x_0|(|y_0|+1)< \frac{\epsilon}{2}$. Combining yields: $|(x-x_0)(y-y_0)| \cdot(|x_0|+1) +|x-x_0|(|y_0|+1) < \epsilon$. I've tried manipulating the left hand side to get $|xy-x_0y_0|$ with little success. Does this whole approach to the problem seem ill-conceived? And if so, how can I rectify it? Or should I try a totally different approach? Thanks.


My approach would have been:



Then noting that $|x|-|x_{0}|\le|x-x_{0}|<1$ (using that $|x-x_{0}|<1$ since it is bounded by the min of $1$ and $\frac{\epsilon}{2(|x_{0}|+1)}$) so $|x|<|x_{0}|+1$. So the above is bounded by


  • $\begingroup$ Why is there an $\leq$ sign towards the end of your second line, instead of just a $<$? $\endgroup$ – James Pirlman Aug 20 '13 at 3:21
  • $\begingroup$ I did that because there is already a strict inequality at the beginning of the second line so in the end I will have the strict inequality I desire anyway. But you are correct that it is in fact strict. I will edit it. $\endgroup$ – user71352 Aug 20 '13 at 3:24
  • $\begingroup$ Great, thanks for the answer, this was a very clear solution. $\endgroup$ – James Pirlman Aug 20 '13 at 3:25
  • $\begingroup$ Your welcome. Hope it helps you understand. $\endgroup$ – user71352 Aug 20 '13 at 3:26
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    $\begingroup$ Yeah, now it's clear to me I should have started out by manipulating $|xy -x_0 * y_0|$ into a more amicable form, instead of trying to manipulate the given inequalities until I got something of the form $|xy - x_0 y_0|$, which is more or less stabbing into the dark. $\endgroup$ – James Pirlman Aug 20 '13 at 3:33

This is a standard trick to add and subtract a term so that you end up via triangle inequality with two quantities that you have bounds for:

The quantity you want to bound can be written

$|xy - x_0 y_0| = |xy - x_0y + x_0y - x_0 y_0 | \leq |xy - x_0 y| + |x_0y - x_0 y_0|$

to get you started. You will also need to do one more step to get a $y_0$ into the first piece.

  • 1
    $\begingroup$ For a geometric interpretation, think about the area added when you go from a rectangle $x_0$ by $y_0$ to a rectangle $x$ by $y$. The L shape added splits into the two pieces above. $\endgroup$ – Ted Shifrin Aug 20 '13 at 3:15

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