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

$|y-y_0| < \min(\frac{|y_0|}{2}, \frac{ \varepsilon |y_0|^2}{2})$


$y \neq 0$ 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.

  • 1
    $\begingroup$ Hint: there is a reason they chose $|y - y_0| < \frac{\epsilon y_0^2}{2}$ instead of just $< \epsilon y_0^2$. $\endgroup$ – Paul Sinclair Aug 14 '15 at 22:46
  • $\begingroup$ Is y just arbitrary, or is $\epsilon$-close to $y_{0}$, or is there some similar closeness condition on y? $\endgroup$ – Sinister Cutlass Aug 14 '15 at 22:48
  • $\begingroup$ @SinisterCutlass There aren't any other conditions or stipulations for the question. It just wants you to use the provided relationships. $\endgroup$ – Shane T Aug 14 '15 at 22:52
  • 1
    $\begingroup$ Ah, I see. Sometimes users are incomplete when they write down questions, attempting to paraphrase them from textbooks, so I wanted to make sure. The solution below looks great. As for why $y\neq 0$, it should be clear from the problem statement itself. The consequence of assuming $y_{0}\neq 0$ and $|y-y_{0}|<min(\frac{|y_{0}|}{2}, \frac{\epsilon |y_{0}|^{2}}{2})$ is that $|y-y_{0}|$ is less than BOTH of the elements one takes the minimum of. Replacing y with 0, it then becomes clear that the assumption precludes this. $\endgroup$ – Sinister Cutlass Aug 14 '15 at 22:58

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.

  • 1
    $\begingroup$ Would you mind providing some insight on what observations you made to come up with your solution? I understand your solution as I read through it, but when I'm doing these problems myself I find it's hard for me to determine how to add all the pieces together. $\endgroup$ – Shane T Aug 14 '15 at 23:11
  • 3
    $\begingroup$ @Shane T Thanks for asking. To large extent, using $\varepsilon-\delta$ language to prove the existence of limit is the combination of algebra simplification and inequality manipulation. Basically you want to control the upper bound of $|1/y - 1/y_0|$, so you want to control the upper bound of the numerator and the lower bound of the denominator, that's how the hint determines that threshold. $\endgroup$ – Zhanxiong Aug 14 '15 at 23:19

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