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I have these statements presented:

$|x - x_0| < \frac{\epsilon}{2(|y_0| + 1)}$ ,

$|x - x_0| < 1$ ,

$|y - y_0| < \frac{\epsilon}{2(|x_0| + 1)}$

And I must prove that:

$|xy - x_0y_0| < \epsilon$

The directions of the problem state: "One word of advice: since the hypotheses only provide information about $x - x_0$ and $y - y_0$, it is almost a foregone conclusion that the proof will depend upon writing $xy - x_0y_0$ in a way that involves $x - x_0$ and $y - y_0$"

My question is - does the solution involve checking every case for $|x - x_0|$ like having same/different signs and the same for $y$ and $y_0$?I'm almost certain there must be a better way, because the way I'm going it's getting me nowhere, I'm sure I'm going at this wrong.I start to do something like:

$x < \frac{\epsilon}{2(|y_0| + 1)} + x_0$

$y < \frac{\epsilon}{2(|x_0| + 1)} + y_0$

But with more cases and it goes completely off.Can someone offer directions?

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  • $\begingroup$ What happens if you multiply $|y-y_0||x-x_0|$? $\endgroup$ – abiessu Nov 13 '13 at 1:50
  • $\begingroup$ @abiessu Shouldn't I be using addition?How will multiplication help?By multiplying them I just get $|xy - yx_0 - xy_0 + x_0y_0|$ , but by adding them I don't get anything sensible either... $\endgroup$ – darkradeon Nov 15 '13 at 21:37
  • $\begingroup$ oh nevermind I got it! I inserted "virtual" variables like this: $|xy - x_0y_0| = |xy - xy_0 + xy_0 - x_0y_0| \leq |xy - xy_0| + |xy_0 - x_0y_0|$ (the triangle inequality... sort of) and now I can just take out multipliers from the 2 right parts to make it into: $|x||y - y_0| + |y_0||x - x_0|$ and this is just the ones in the problem conditions, so plugging them into the original expressions and combining them: $|x||y - y_0| + |y_0||x - x_0| < \frac{|x|\epsilon}{2(|x_0| + 1)} + \frac{|y_0|\epsilon}{2(|y_0| + 1)}$ and since $\frac{|y_0|\epsilon}{2(|y_0| + 1)}$ is always positive, removing it... $\endgroup$ – darkradeon Nov 15 '13 at 22:33
  • $\begingroup$ Interestingly, your solution comes up with the same three $\epsilon$ terms as appear in the answer I came up with. In particular, note that $|x|\le |x-x_0|+|x_0|$... $\endgroup$ – abiessu Nov 17 '13 at 14:54
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Considering the quantity $|y-y_0|\cdot|x-x_0|,$ we get

$$|xy-xy_0-x_0y+x_0y_0|\lt \frac {\epsilon^2}{4(|x_0|+1)(|y_0|+1)}\tag 1$$

Adding $|y_0(x-x_0)|$ and $|x_0(y-y_0)|$ to the left-hand side of $(1)$, we get

$$|xy-xy_0-x_0y+x_0y_0|+|yx_0-y_0x_0|+|xy_0-x_0y_0|\ge |xy-x_0y_0|\tag 2$$

$(2)$ is due to the triangle inequality. Note that the left-hand side of $(2)$ obeys

$$|xy-xy_0-x_0y+x_0y_0|+|yx_0-y_0x_0|+|xy_0-x_0y_0|\lt \frac {\epsilon}{2(|x_0|+1)}+\frac {\epsilon|x_0|}{2(|x_0|+1)}+\frac {\epsilon|y_0|}{2(|y_0|+1)}\tag 3$$

(from the given inequality $|x-x_0|\lt 1$.) Therefore, putting $(2)$ and $(3)$ together we get

$$\frac {\epsilon}{2(|x_0|+1)}+\frac {\epsilon|x_0|}{2(|x_0|+1)}+\frac {\epsilon|y_0|}{2(|y_0|+1)}\gt |xy-x_0y_0|\tag 4$$

and the LHS of $(4)$ becomes

$$\epsilon\frac {|y_0|+1+|x_0|(|y_0|+1)+|y_0|(|x_0|+1)}{2(|x_0|+1)(|y_0|+1)}\le \epsilon$$

with the final result

$$|xy-x_0y_0|\lt \epsilon$$

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