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