Let us take the most basic example:
$\lim_{n\to\infty} \frac{n!}{n^n}$
$\frac{n!}{n^n} = \frac{n(n-1)...1}{n \cdot n...n} = \frac{n}{n} \cdot \frac{n-1}{n} \cdot [...] \cdot \frac{1}{n} \leq \frac{1}{n}$ We used the fact that each term was smaller or equal than 1, so the result is upper-bounded by $\frac{1}{n}$, and its limit goes to 0.
This is all fine and good; the question is whether we are allowed to use this trick within a limit. In the above example, we've shown the inequality for $n \in \mathbb{N}$, and then we took its limit. Am I right in saying that it would be wrong to write: $\lim_{n\to\infty}\frac{n!}{n^n} = \lim_{n\to\infty}\frac{n(n-1)...1}{n \cdot n...n} = \lim_{n\to\infty}\frac{n}{n} \cdot \frac{n-1}{n} \cdot [...] \cdot \frac{1}{n} \leq \lim_{n\to\infty}\frac{1}{n}$ ?
In this second case, there are actually an infinite amount of terms $\frac{i}{n}$ for $ i \in \mathbb{N}$, therefore when need to take the limit of each term (infinitely many). Intuitively, it's clear why this would still 'work' in this example because each term has a limit, but to me it seems that this isn't formal... In the same sense that we cannot deal with infinite dimensional vector spaces the way we can with finite ones, is there a similar difference between expressions containing uncountably infinitely many terms and those containing countably many?