calculus
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 $$ \lim_{x\to \dfrac{1}{\pi}} \dfrac{\pi}{x}=\pi^2$$
By the mean value theorem $|f(x)-L|=|x-c|\cdot f'(\xi)$ for some $\xi$ between $x$ and $c$. If $\epsilon, \delta$ are small then $f'(\xi)\approx f'(c)$ for all $\xi\in (c-\delta,c+\delta)$ and hence $|x-c|<\frac \epsilon{|f'(c)|}$ implies $|f(x)-L|\stackrel <\approx \epsilon $.
Or to put it differently: Near $c$ the function $f(x)$ approximately looks like $g(x)=f(c)+(x-c)f'(c)$ and for $g$ it is clear that $\delta=\frac \epsilon{|f'(c)|}$ is the best choice.
Let's make the above rigorous:
We are given an open interval $I\subseteq \mathbb R$ and a continuoiusly differentiable function $f\colon I\to \mathbb R$ and a point $c\in I$ and with $f'(c)\ne 0$.
Consider the set $$\mathcal D = \{h\colon (0,\infty)\to(0,\infty)\mid\forall\epsilon>0\colon\forall x\in I\colon |x-c|<h(\epsilon)\Rightarrow |f(x)-f(c)|<\epsilon\}.$$ Then $\mathcal D$ contains all functions that one is allowed to use for finding a suitable $\delta$, given $\epsilon$ in determining $\lim_{x\to c}f(x)$. Unfortunately, there is no natural order relation on $\mathcal D$, hence it is not immediatel clear how to apply terms like "least upper bound" to $\mathcal D$. However, we have a partial order (not only on $\mathcal D$, but on the set of all functions $(0,\infty)\to(0,\infty)$) as we can define $h_1\le h_2$ if $\limsup_{x\to 0}\frac{h_1(x)}{h_2(x)}\le 1$.
Let $H(x)=\frac x{|f'(c)|}$ and $\tilde h(\epsilon)=\inf\{|x-c|\mid |f(x)-f(c)|\ge\epsilon\}$. Note that in general $H\notin \mathcal D$. We prove the following statements:
These three statements together allow us to loosely speak of $H(\epsilon)$ being an approximation to the maximal possible choice of $\delta$ for sufficiently small $\epsilon$.
Lemma 1: $\tilde h\in\mathcal D$.
Proof: Since $f$ is continuous, for $\epsilon>0$ there exists $\delta>0$ such that $|x-c|<\delta$ implies $|f(x)-f(c)|<\epsilon$. We conclude that $\tilde h(\epsilon)\ge \delta>0$, hence $h$ is indeed a function $(0,\infty)\to (0,\infty)$. If $\epsilon>0$ is given and $|x-c|<\tilde h(\epsilon)$, then $|f(x)-f(c)|<\epsilon$, as otherwise we would have $\tilde h(\epsilon)\le |x-c|$. Therefore $\tilde h \in \mathcal D$.$_\blacksquare$
Lemma 2: If $h\in \mathcal D$ then $H\ge h$.
Proof: We have to show that for $\epsilon>0$ there exists $\delta>0$ such that $0<x<\delta$ implies $\frac{H(x)}{h(x)}>1-\epsilon$, i.e. $H(x)>h(x)(1-\epsilon)$. I am a bit too tired for these calculations right now, so I will add them tomorrow - one has to be careful not to mix $\epsilon, \delta$ belonging to the $\limsup$ with those belonging to $h$ or the continuity of $f$.
Lemma 3: $\tilde h\ge H$.
Proof: Same here, see you tomorrow.
I think your main issue is that you are still trying to think about this exercise as a routine algebraic manipulation. It is not exactly like that.
The thing is that here we have a goal/target about ensuring that some inequality holds. In current question the goal is to ensure that $$|x^2-9|<\epsilon$$ We are not supposed to find all values of $x$ for which the above inequality holds (similar to solving equations like $x^2=9$). The problem is not exactly algebraic. Rather what we desire is to find a range of values of $x$ near $3$ for which this inequality can be ensured. Such a range of values of $x$ may or may not exist. Our task is to prove that such a range of values of $x$ near $3$ always exists no matter what $\epsilon $ is given.
The technique is to replace the target inequality by a simpler one. Thus we have to find some expression $g(x) $ which is simpler in form and satisfies $$|x^2-9|<g(x)$$ and then replace the goal with ensuring that $g(x) <\epsilon $. Thus our original target is to be achieved via a combination of two simpler goals $|x^2-9|<g(x)$ and $g(x) <\epsilon$.
The problem is now to choose a suitable $g(x) $ and to find a range of values of $x$ near $3$ which can ensure that both the sub-goals are met. This is where one has great leverage and problem is actually far simpler than it appears. We have $$|x^2-9|=|x+3||x-3|$$ Now let us choose any specific range of values of $x$ near $3$, say $|x-3|<1$ (this is totally as per your wish, but in general the range should be such that the desired simplification in what follows is possible). And $$|x+3|\leq |x-3|+6<7$$ and therefore we have $$|x^2-9|=|x+3||x-3|<7|x-3|$$ for the range of values of $x$ given by $|x-3|<1$.
Thus we can choose $g(x) =7|x-3|$ and one of the sub goals is achieved for the range $|x-3|<1$. The other goal is now simpler $$7|x-3|<\epsilon $$ Obviously this can be achieved by the range of values of $x$ given by $|x-3|<\epsilon /7$ (if this is not obvious to you then you need to see how inequalities work in general).
So for the two goals we have found two ranges of values of $x$ namely $|x-3|<1$ and $|x-3|<\epsilon /7$ which ensures that the respective goals are met. Since we want to ensure that both the goals are met simultaneously we need to deal with the range of values of $x$ which are common to both $|x-3|<1$ and $|x-3|<\epsilon/7$. This is possible if $|x-3|<\min(1,\epsilon /7)$ and we are done by setting $\delta=\min(1,\epsilon/7)$ and our desired range of values of $x$ is $|x-3|<\delta$.
The important thing to notice here is that our original problem to ensure some inequality is replaced by two far simpler (but not necessarily equivalent) problems. This is in quite contrast to solving equations like $x^2-9=0$ where the problem is reduced to two simpler and equivalent problems $x-3=0,x+3=0$.
The fact that we have to simplify the problem without caring about equivalence gives us great leverage here. Most beginners however don't notice this and instead focus on solving inequalities (where the problem can be simplified but only to an equivalent one) and this is one of stumbling blocks in understanding and applying definition of limit.
More formally the target inequality $$|f(x) - L|<\epsilon $$ is not a hypothesis but a conclusion in a long chain of logical implications. Also by definition the implications involved are one way and you don't need to put any extra effort to unnecessarily ensure a both way implication. And we present our argument like "the target conclusion, say $A$, holds if (not iff) $B, C, \dots$ hold and so on till we reach a stage where we get to see ranges of values of $x$". So the chain of implications is figured out in reverse.
Using your own words from question: how $$|x+3||x-3|<\epsilon$$ and $$|x+3||x-3|<C|x-3|$$ lead to $$|x-3|<\epsilon /C$$ is not the right question, but you should ask how $$|x-3|<\epsilon /C$$ and $$|x+3||x-3|<C|x-3|$$ lead to $$|x+3||x-3|<\epsilon $$ This is the desired logical flow and it would now look obvious to you. The thing however is that the individual logical implications have to be figured out in reverse starting from the conclusion to the hypotheses.
Years of training in algebraic manipulation which are mostly forward or both way implications makes things in analysis a little bit surprising (if not difficult) when we have to deal with one way implications in reverse manner. Thus we switch from "$A$ implies $B$" to "$B$ holds if $A$ holds".
Best Answer
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 $\delta := \min \left\{ \frac{1}{2\pi}, \varepsilon/2\pi^{3} \right\}$.