Letting $\varepsilon \to 0$ in proofs

Question

I've seen proofs regarding supremum and infimum of bounded sets in $\mathbb{R}$ involving an arbitrary $\varepsilon > 0$ and 'hard' inequalities (not necessarily strict, but hard to manipulate algebraically and wrt ordering). Most of what I saw involves the argument "letting $\varepsilon \to 0$" and I'm kind of baffled. Why is this logically valid?

For context, consider this proof showing that $\sup(\frac{1}{A}) = \frac{1}{\inf A}, \inf A > 0$ (Please check if the steps are correct).

Proof. $\sup(\frac{1}{A}) \leq \frac{1}{\inf A}$ is easy to show. Now for $\sup(\frac{1}{A}) \geq \frac{1}{\inf A}$,

Let $\varepsilon > 0$. Then $\exists x_{\varepsilon} \in A$ such that $\inf A + \varepsilon > x_{\varepsilon}$. Then $\frac{1}{\inf A + \varepsilon} < \frac{1}{x_{\varepsilon}} \leq \sup(\frac{1}{A})$. Letting $\varepsilon \to 0$, $\frac{1}{\inf A} \leq \sup(\frac{1}{A})$.

Answer 1

You have proven the following:

$$\forall \epsilon > 0: \frac{1}{\inf A + \epsilon} \le \sup\left(\frac{1}{A}\right)$$

Now, appeal to the following: if $f(x) \le g(x)$ for all $x$ in a neighborhood of $a$, then $$\lim_{x \to a}f(x) \le \lim_{x \to a}g(x)$$ (provided the limits exist).

Taking $f(\epsilon) = 1/(\inf A + \epsilon)$ and $g(\epsilon) = \sup(1/A)$ (constant function), you get the desired result.

Answer 2

That works because if a function $f$ is such that $f(x)\leqslant M$ for some constant $M$ and if $\lim_{x\to a}f(x)$ exists, then $\lim_{x\to a}f(x)\leqslant M$. This is so because, if $l=\lim_{x\to a}f(x)$ and if $l>M$, then there is some $\delta>0$ such that $|x-a|<\delta\implies\left|f(x)-l\right|<l-M$. But then, if $|x-a|<\delta$,$$l-f(x)=-\bigl(f(x)-l\bigr)\leqslant|f(x)-l|<l-M,$$ and therefore $f(x)>M$. But we are supposing that we always have $f(x)\leqslant M$.

So, in that situation that you have described, since$$(\forall\varepsilon>0):\frac1{\inf(A)+\varepsilon}\leqslant\sup\left(\frac1A\right),$$then$$\frac1{\inf(A)}=\lim_{\varepsilon\to0}\frac1{\inf(A)+\varepsilon}\leqslant\sup\left(\frac1A\right).$$