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Prove that if $|x-x_0| < \min\left(\frac{\epsilon}{2(|y_0| + 1)}, 1\right)$ and $|y-y_0| < \frac{\epsilon}{2(|x_0| + 1)},$ then $|xy - x_0y_0| < \epsilon.$

I am doing some problems in Spivak's Calculus on inequalities and came across this problem. Currently I have a sketch of a solution that breaks down the problem into many cases and it is kind of long and messy. I thought maybe someone here could provide a clean and easier solution? If there is a nice solution please tell me a bit behind the thought process (like how you came up with it), instead of giving it as it is.

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    I don't $k$$n$ow if this is too obvious, but it see$m$s like a delta-epsilon argument to show that f(x,y)=xy is continuous at (xo,yo). Look at 2 possible cases:2011-04-14

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$\begin{align*} |xy-x_0y_0| &= |xy-xy_0 + xy_0 - x_0y_0|\\ &= |x(y-y_0) + (x-x_0)y_0|\\ &\leq |x(y-y_0)| + |y_0||x-x_0|\\ &= |(x-x_0)(y-y_0) + x_0(y-y_0)| + |y_0||x-x_0|\\ &\leq |x-x_0||y-y_0| + |x_0||y-y_0| + |y_0||x-x_0|\\ &\leq |y-y_0| + |x_0||y-y_0| + |y_0||x-x_0| \\ &= (1+|x_0|)|y-y_0| + |y_0||x-x_0|\\ &\lt (1+|x_0|)\left(\frac{\epsilon}{2(|x_0|+1)}\right) + |y_0|\left(\frac{\epsilon}{2(|y_0|+1)}\right)\\ &= \frac{\epsilon}{2} +\left(\frac{|y_0|}{|y_0|+1}\right)\left(\frac{\epsilon}{2}\right)\\ &\lt \frac{\epsilon}{2} + \frac{\epsilon}{2}\\ &= \epsilon. \end{align*}$ We can do the step in line $6$, because $|x-x_0|\lt 1$; we can do the step in line $8$ because $|x-x_0|\lt \frac{\epsilon}{2(|y_0|+1)}$ and $|y-y_0|\leq\frac{\epsilon}{2(|x_0|+1)}$. We can do the step in line 10 because $\frac{|y_0|}{|y_0|+1}\lt 1$.

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    looks right to me, I definitely need more practice on inequalities.2011-04-14
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Here is another slightly less familiar trick. $xy-x_0y_0=(x-x_0)(y-y_0)+y_0(x-x_0)+x_0(y-y_0)$

Natural, I think, for this problem, since we have some information about $|x-x_0|$ and $|y-y_0|$ in terms of $y_0$ and $x_0$.

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    @Glen, @user6312: fixed. Thanks, Glen, for flagging this.2011-04-14