So I've asked this same question in the past but I'm still troubled with this problem. They have simply asked for a proof of this limit using the delta epsilon definition of limit. I have a proof which I can follow but the thing which I can't quite get is how they choose $\delta=\min\left(\dfrac{1}{2\pi},\dfrac{\epsilon}{2\pi^3}\right)$. I'm hoping that someone can explain the process behind finding these $\delta$

The limit is $$ \lim_{x\to \dfrac{1}{\pi}} \dfrac{\pi}{x}=\pi^2$$

  • $\begingroup$ You should give a reference to the proof or reconstruct it, either it is not so easy to explain. Normally you choose the $\delta$ such that the inequality holds. Therefore you never choose it at the begin, it is written, but at that part, where you finish your proof. $\endgroup$ – Mundron Schmidt Jun 9 '17 at 16:55
  • $\begingroup$ well I don't need a specific solution to this problem but what I'm looking for is a way to think about choosing a $\delta$ in these sort of cases. I get the logic behind the linear, quadratic, and square root cases, but not this type. $\endgroup$ – john fowles Jun 9 '17 at 17:00

If $x \neq 0$, then $$ \left|\frac{\pi}{x} - \pi^{2}\right| = \pi \left| \frac{1}{x} - \pi\right| = \pi^{2}\left|\frac{x - \frac{1}{\pi}}{x}\right|. $$ If, in addition, we have $\left|x - \frac{1}{\pi}\right| < \frac{1}{2\pi}$ (this is to bound away the annoying denominator by preliminarily bounding $\left|x-\frac{1}{\pi}\right|$), then $\frac{1}{2\pi} < x < \frac{3}{2\pi}$, and hence $$ \pi^{2}\left| \frac{x - \frac{1}{\pi}}{x}\right| < \pi^{2}\cdot 2\pi \left|x - \frac{1}{\pi}\right| = 2\pi^{3}\left|x - \frac{1}{\pi}\right|. $$ Given any $\varepsilon > 0$, we have $2\pi^{3}\left|x - \frac{1}{\pi}\right| < \varepsilon$ if further we have $|x - \frac{1}{\pi}| < \varepsilon/2\pi^{3}$. To make all the above "if"'s happen simultaneously, it suffices to take $\delta := \min \left\{ \frac{1}{2\pi}, \varepsilon/2\pi^{3} \right\}$.

  • $\begingroup$ Everything you wrote is great and nearly the same as the proof that I have but could you explain why you specifically chose $\dfrac{1}{2\pi}$. What was your procedure to find this? I know that it's sufficient in these proofs to show that $|f(x)-L|<|x-c|$ so I was looking at comparing $\dfrac{\pi^2}{|x|}<1$ and finding the minimum value of $x$ when it's close to $\dfrac{1}{\pi}$. I'm not sure if i'm thinking in the right direction or if you chose $\dfrac{1}{2\pi}$ for some other reason $\endgroup$ – john fowles Jun 9 '17 at 17:12
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    $\begingroup$ The choice is just about convenience and the inequality :D. You can use any real number $r > 0$ such that $\frac{1}{\pi} - r> 0$, so that you get $ \frac{1}{\pi} - r < |x|$ and then the last term in the first independent line can get larger you know. It is actually very simple (though it would seem abstruse at the first glance). $\endgroup$ – Megadeth Jun 9 '17 at 17:16
  • $\begingroup$ This comment is great! I really over complicated the selection of $\delta=\dfrac{1}{2\pi}$. This clears it up $\endgroup$ – john fowles Jun 9 '17 at 17:47

Let be $\varepsilon>0$. Then choose $\delta= \ldots$.

First we do not choose $\delta$, but we leave it open...

Let be $x\in\mathbb{R}$ such that $\left|x-\frac1{\pi}\right|<\delta$. We get $$ \left|\frac{\pi}x-\pi^2\right|=\left|\pi^2\left(\frac{\frac1\pi-x}x\right)\right|=\pi^2\frac{\left|\frac1\pi-x\right|}{|x|}<\pi^2\frac{\delta}{|x|}. $$ Now you try to achieve $<\varepsilon$ at the end. For this purpose you can use, that $\delta$ isn't chosen yet. You have a lot of options! One is to say, that you claim $\delta<\frac{\varepsilon}{2\pi^3}$. In that case you get $$ \pi^2\frac{\delta}{|x|}<\frac{\varepsilon}{2\pi|x|}. $$ Now you have to eliminate $|x|$. From $\left|x-\frac1\pi\right|<\delta$ you get $|x|>\frac1\pi-\delta$. To finish the proof you claim $\delta<\frac1{2\pi}$ such that you get $|x|>\frac1{2\pi}$. Now follows $$ \frac{\varepsilon}{2\pi|x|}<\varepsilon. $$ Your proof is finished and you claimed $\delta<\frac1{2\pi}$ and $\delta<\frac{\varepsilon}{2\pi^3}$. Therefore you have to plug in $\delta<\min\left\{\frac1{2\pi},\frac{\varepsilon}{2\pi^3}\right\}$ at the beginning.


Since $x$ is near to $\frac1\pi$ you can assume, that $|x|$ is also near to $\frac1\pi$. Therefore you can see $$ \pi^2\frac{\delta}{|x|}\sim\pi^3\delta. $$ That gives the idea, that $\delta$ should be at least less than $\frac{\varepsilon}{pi^3}$. To be save, we claim $\delta$ to be even smaller, namely smaller than $\frac{\varepsilon}{2\pi^3}$. The second choice follows then directly, such that the remaining constants vanish.


In a comment the OP states:

...but what I'm looking for is a way to think about choosing a δ in these sort of cases.

Here is 'rough' (contains sloppy logic and pitfalls) equivalence chain for the general case:

$\quad \lim_{x\to a} f(x) = f(a) \; \text{ iff }$

$\quad |f(x) = f(a)| \lt \varepsilon \; \text{ iff }$

$\quad -\varepsilon \lt f(x) - f(a) \lt +\varepsilon \; \text{ iff }$

$\quad f(a) -\varepsilon \lt f(x) \lt f(a) +\varepsilon \; \text{ iff }$

$\quad f^{-1}(f(a) -\varepsilon) \lt x \lt f^{-1}(f(a) +\varepsilon) \; \text{ iff (are you kidding?)}$

$\tag 1 f^{-1}(f(a) -\varepsilon) -a \le -\delta \lt x - a \le +\delta \lt f^{-1}(f(a) +\varepsilon) -a\; \text{( lay in a delta)}$

There are two problems with the above rough conceptual presentation:

$\quad f \; \text{doesn't have to be an invertible function}$

$\quad \text{Applying 'this' } f^{-1} \text{might 'flip' our inequalities}$

For the OP's problem, considering that $x = 0$ is not in the domain of $f(x) = \frac{\pi}{x}$, we know right off the bat that there must be restrictions on $\delta$.

Our function is certainly invertible on the open interval

$\quad (0, +\infty)$

so we are only interested in what happens there.

When you consider how to nest in a $\delta$ that works with a properly analyzed $\text{(1)}$, you might come up with the following observation about our decreasing function $f(x)$:

For $a \gt 0$ and $0 \lt \delta \lt a$,
$\tag 2 f(a) - f(a + \delta) \lt f(a - \delta) - f(a)$

This is a simple thing to prove using algebra.

So to 'control' our function $f(x)$ with a $\delta$ we only have to work on the left side of $a = \frac{1}{\pi}$.

Now if we want

$\quad f(\frac{1}{\pi} - \hat{\delta}) - f(\frac{1}{\pi}) \lt \varepsilon$


$\quad \frac{\pi}{\frac{1}{\pi} - \hat{\delta}} - \pi^2 \lt \varepsilon$

After some algebra we conclude that

$\delta = \frac{\varepsilon}{\pi^3 + \pi \varepsilon}$


Notice that this number $\delta$ is less than $\frac{1}{\pi}$, no matter what number $\varepsilon \gt 0$ is given.

Note: This might look a bit crazy buy I've be thinking about this 'turn-the-crank' approach to epsilon/delta problems for a couple of days. The fewer 'tricks' the better. A robot might be able to do this in the future.


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