I've tried a lot of things, but I can't seem to get anywhere with this problem. I'm hoping the solution is simple but that I'm just missing it.
The problem is as follows:
Prove that if $y_0 \neq 0$ and
$|y-y_0| < \min(\frac{|y_0|}{2}, \frac{ \varepsilon |y_0|^2}{2})$
then
$y \neq 0$ and
$|\frac{1}{y} – \frac{1}{y_0}| < \varepsilon$
I've tried a lot of various things, but nothing seems to get me really any closer.
One thing I think may be related is that you can rewrite
$|\dfrac{1}{y} – \dfrac{1}{y_0}|$ as $|\dfrac{y_0 – y}{y y_0}|$
and from
$|y-y_0| < \dfrac{ \varepsilon |y_0|^2}{2}$ you can say that
$|\dfrac{y-y_0}{y_0^2}| < \dfrac{ \varepsilon}{2}$ which implies that
$|\dfrac{y-y_0}{y_0^2}| < \varepsilon$
So if you can prove that
$|\dfrac{y-y_0}{y y_0}| \leq |\dfrac{y-y_0}{y_0^2}|$ (which may not even be true) then you can prove that
$|\dfrac{y-y_0}{y y_0}| = |\dfrac{1}{y} – \dfrac{1}{y_0}| < \varepsilon$
which all hinges on proving that $|y| \leq |y_0|$, which may not even be true, which would make all my assumptions fall apart.
Also, as far as proving that $y \neq 0$,
I can easily prove that $y \neq 0$ from the first part of the inequality $|y-y_0| < \frac{y_0}{2}$, but I have no idea how to prove it from the other statement that $|y-y_0| < \frac{ \varepsilon |y_0|^2}{2}$.
In any case, I'm convinced I'm going about this the wrong way, and am missing some key observation that would allow me to solve this. Thanks for any help or insight you're able to give.
Best Answer
By triangle inequality $$\left|y_0\right| = \left|(y_0 - y) + y\right| \leq \left|y - y_0\right| + \left|y\right|,$$ $\left|y - y_0\right| < \dfrac{\left|y_0\right|}{2}$ implies that $$\left|y\right| \geq \left|y_0\right| - \left|y - y_0\right| > \left|y_0\right| - \frac{\left|y_0\right|}{2} = \frac{\left|y_0\right|}{2}.$$
Now, apply the condition $\left|y - y_0\right| < \frac{\varepsilon}{2}\left|y_0\right|^2$ to the numerator in the following expression: $$\left|\frac{1}{y} - \frac{1}{y_0}\right| = \frac{\left|y - y_0\right|}{\left|y\right|\left|y_0\right|} < \frac{\left|y - y_0\right|}{\frac{\left|y_0\right|}{2}\left|y_0\right|} < \frac{\frac{\varepsilon}{2}\left|y_0\right|^2}{\frac{\left|y_0\right|^2}{2}} = \varepsilon$$ hence the result follows.