# finding upper bound for delta epsilon definition of limit proof

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

• 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. – Mundron Schmidt Jun 9 '17 at 16:55
• 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. – 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 $\delta := \min \left\{ \frac{1}{2\pi}, \varepsilon/2\pi^{3} \right\}$.

• 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 – john fowles Jun 9 '17 at 17:12
• 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). – Megadeth Jun 9 '17 at 17:16
• This comment is great! I really over complicated the selection of $\delta=\dfrac{1}{2\pi}$. This clears it up – 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.

Remark:

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

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

Then

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

After some algebra we conclude that

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

works.

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.