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Supposing that $|x-x_0| < min(1, \frac{\epsilon}{2(|y_0|+1)})$ and $|y-y_0| < \frac{\epsilon}{2(|x_0|+1)}$ it follows that $|{xy-x_0y_0}| < \epsilon$

(A friend told me it was helpful to prove this in hopes of proving that $\lim_{x \to c}(f*g)=L*M$ where $\lim_{x \to c}f(x) = L$ and $\lim_{x \to c}g(x) = M$)

I think it's supposed to end something like $|{xy-x_0y_0}| <|x-x_0||y-y_0|<(2(|x_0|+1))*\frac{\epsilon}{2(|x_0|+1)}<\epsilon$ and you're finished, but I get stuck in the steps before.

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See… – wj32 Oct 30 '12 at 22:50
I think that proves what I'm trying to prove with this, but not this directly. Also, it uses limits, but not with epsilon delta definitions and what not, which, sorry for not mention, was what I intended the proof to use, or at least I think it does. – Casquibaldo Oct 30 '12 at 23:03

Hint $$|xy-x_0y_0|=|xy-xy_0+xy_0-x_0y_0| \leq |x| |y-y_0|+|y_0| |x-x_0|$$

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I think I can work it out from here. – Casquibaldo Oct 30 '12 at 23:07
@Casquibaldo This addition of cross term is a pretty standard technique, once you see it once you will be probably able to use it again in other contexts... – N. S. Oct 30 '12 at 23:12
Yeah, it seems like one of those things that you must have seen before to know that it is appropriate to use. Still a good technique though. Also, wouldn't the last $|y|$ that appears in the inequality be a $|y_0|$? @N.S. – Casquibaldo Oct 30 '12 at 23:17
@Casquibaldo yup, fixed that. ty – N. S. Oct 31 '12 at 3:03
I am sorry: please could you work out full details? I tried but could not do it! – blue Dec 20 '13 at 8:20

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