$\varepsilon$-$\delta$ proofs seem to be the most confusing concept of first-year Calculus (or pre-Calculus, depending on how it's taught).

The $\varepsilon$-$\delta$ definition of a limit is as follows. $\lim_{x\rightarrow a} f(x) = l$ means for all $\varepsilon > 0$, there is a $\delta > 0$ such that if $|x - a| < \delta$ then $|f(x) - l| < \varepsilon$. All your answers are contained within this definition so let's break it down and see how all the proofs you reference satisfy the definition.

For the limit to be $l$, we require that every $\epsilon > 0$ has a corresponding $\delta > 0$. A standard way to prove that something holds for all values of a variable in a certain domain is to choose an arbitrary (unspecified) value for that variable in its domain and make a proof that works for that arbitrary choice. Spivak's proof does exactly that: he says "for any $\varepsilon > 0$" and then proceeds to show that there is a $\delta$ corresponding to this specific but arbitrary $\varepsilon$. The unstated conclusion is that there is a $\delta$ for every $\varepsilon$.

Note that the proof of a limit requires that there is a $\delta$ corresponding to every positive "epsilon", but there is no requirement to denote the "epsilon" value as the symbol $\varepsilon$. Instead, we may use an expression such as $c\varepsilon$ for constant $c>0$ so long as this expression can take on all values in the domain of the "epsilon" (namely, all positive numbers). Clearly we may write any positive number as $c\varepsilon$ for some choice of $\varepsilon$. So if we prove that a $\delta$ exists for all $c\varepsilon$ in place of $\varepsilon$, the proof holds for every value that $c\varepsilon$ can take, which means it holds for every positive number, so we are good.

However, if the expression we choose *can't* take on all possible "epsilon" values, such as $1+\varepsilon$ for $\varepsilon > 0$, then if we prove that a $\delta$ exists for all $1+\varepsilon$ then we *haven't* proven that a $\delta$ exists for all values for which the definition of limit requires it exist. Such a proof would be invalid.

Finally, you posted a link to another question where a complex expression $\min\left(1,\frac{\epsilon}{2(|m|+1)}\right)$ stands in for the "epsilon". Note that by letting $\epsilon$ vary, this expression can take on all positive values up to but not exceeding 1. Therefore it *doesn't* take on all values that the definition of limit requires. However, if a $\delta$ works for some "epsilon" then it automatically works for any larger "epsilon" - you can immediately verify this statement in the definition of a limit. So if our expression can take on all positive numbers up to 1, that's actually good enough. (In the most generality, we can prove that there is a $\delta$ corresponding to all "epsilon" in a set of positive numbers containing a sequence going to 0.)

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

This isn't an $\epsilon - \delta$ proof but rather a standard sequence convergence proof. Further, the limit exists for all $x>x_0$ so I'm not sure why that is part of the question. What you are really trying to prove is that for any positive number $a$ we have

$$\lim_{n\to\infty}a^{1/n}=1$$

There are many ways to prove this but I prefer this way: We have that

$$a^{1/n}=e^{\ln(a^{1/n})}=e^{\frac{\ln(a)}{n}}$$

Since $\ln(a)$ is a real number and $e^x$ is continuous at $x=0$ we have

$$\lim_{n\to\infty}a^{1/n}=\lim_{n\to\infty}e^{\frac{\ln(a)}{n}}=e^{\lim_{n\to\infty}\frac{\ln(a)}{n}}=e^0=1$$

Of course, this proof might not suffice for you depending on what you have already proved but its a good start nonetheless.