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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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    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.2012-10-30

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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 am sorry: please could you work out full details? I tried but could not do it!2013-12-20