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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 haven't thought through this at all, but maybe one of those tricks like rewriting $xy - x_0 y_0$ as $xy - x_0 y + x_0 y - x_0 y_0$ helps. Of course, if it does, this doesn't necessarily help understand, since it would just be a little algebraic trick. Though I guess there's value in learning this trick as a tool. –  matt Apr 14 '11 at 4:08
    
@matt: That was my first impression on the problem as well: "hmm maybe some algebraic trick would do the job." but it turned out to be long and messy. I guess the key is which algebraic trick? –  Mark Apr 14 '11 at 4:11
    
I don't know if this is too obvious, but it seems like a delta-epsilon argument to show that f(x,y)=xy is continuous at (xo,yo). Look at 2 possible cases: –  gary Apr 14 '11 at 5:08

2 Answers 2

up vote 4 down vote accepted

$$\begin{align*} |xy-x_0y_0| &= |xy-xy_0 + xy_0 - x_0y_0|\\ &= |x(y-y_0) + (x-x_0)y_0|\\ &\leq |x(y-y_0)| + |y_0||x-x_0|\\ &= |(x-x_0)(y-y_0) + x_0(y-y_0)| + |y_0||x-x_0|\\ &\leq |x-x_0||y-y_0| + |x_0||y-y_0| + |y_0||x-x_0|\\ &\leq |y-y_0| + |x_0||y-y_0| + |y_0||x-x_0| \\ &= (1+|x_0|)|y-y_0| + |y_0||x-x_0|\\ &\lt (1+|x_0|)\left(\frac{\epsilon}{2(|x_0|+1)}\right) + |y_0|\left(\frac{\epsilon}{2(|y_0|+1)}\right)\\ &= \frac{\epsilon}{2} +\left(\frac{|y_0|}{|y_0|+1}\right)\left(\frac{\epsilon}{2}\right)\\ &\lt \frac{\epsilon}{2} + \frac{\epsilon}{2}\\ &= \epsilon. \end{align*}$$ 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. –  Mark Apr 14 '11 at 5:29

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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could you elaborate a little more? –  Mark Apr 14 '11 at 4:44
    
The "standard" identity works poorly because, for example in the $y(x-x_0)$ part, we don't have easy control over $y$. In the identity that I mention, we have for example the second term $y_0(x-x_0)$, which by the given information has absolute value less than $\epsilon|y_0|/(2|y_0+1)$, so at least less than $\epsilon/2$, which is more or less what we want. The third term has essentially the same nature, and the first term $(x-x_0)(y-y_0)$ is "very" small. –  André Nicolas Apr 14 '11 at 5:23
    
Help! I cannot find the edit button to fix the missing $ sign. –  André Nicolas Apr 14 '11 at 5:28
    
@user6312 I've flagged the comment, so hopefully a mod will come and save it for you :). –  Glen Wheeler Apr 14 '11 at 7:26
    
@Glen: Thanks, it is embarrassing but not important. I just wanted to provide additional motivation for the trick. –  André Nicolas Apr 14 '11 at 7:43

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