Q:

Is in general incorrect taking limits of inequalities?

I am confused about when the use of limits as a possible manipulation of a given inequality could be considered correct. Below there are some examples and the questions are at the end (this doubt arose from this other previous question):

For instance in $(1,\infty]$ is true that:

$$\frac{1}{n} \lt n$$
Applying $\lim_{n\to\infty}$ to both terms:
$$\lim_{n\to\infty}\frac{1}{n} \lt \lim_{n\to\infty}n$$
It is still true because $0 \lt \infty$. 

But in general it is not possible to use it, for instance in this example kindly provided by @MathMajor in this other question:

$$\frac{1}{n} > 0 \implies \lim_{n \to \infty} \frac{1}{n} > \lim_{n \to \infty} 0 \implies 0 > 0.$$

And finally, in the other hand for instance you could have an inequality that indeed is defined from the very beginning including limits, so I understand that those inequalities are already correct (like this quite trivial example):

$$\lim_{y\to\infty} \frac{x}{y} \lt \lim_{x\to\infty} \frac{x}{y} $$

Is it possible then taking the limit of an inequality or not? Are there cases in which is possible/correct applying limits to the terms of inequalities? Is it possible to apply them adding restrictions? Thank you!

A:

In general, if $a_n \leq b_n $, then $\lim a_n \leq \lim b_n $. Why?
Check: Let $A = \lim a_n $ and $B = \lim b_n$. If, on the contrary, $A > B$, then $A-B>0$. Given: $A = \lim a_n$. This, mean that for every $\epsilon > 0$ (Take $\epsilon = \frac{ A- B}{2} $, for instance), then can find some $N$ so that
$$ |a_n - A | < \epsilon $$
for all $n > N$
Now, with our choice of $\epsilon $, we write 
$$ A - \epsilon < a_n < A + \epsilon \implies \frac{A+B}{2} <a_n $$
We do the same for $\lim b_n =B$ (fill in details ) to obtain 
$$ b_n < \frac{ A + B }{2} $$
and therefore
$$ a_n > b_n \; \; \; for \; \; some \; \; n > N_0 $$
Contradiction, and thus, our claim better be true.

A:

The rule is very simple. Assuming that both limits exist, the inequality is preserved, but only it the "less than or equal" form. Like when you have 
$$
0<\frac1n
$$
for all $n$, but on the limit are equal. In symbols, if $x_n<y_n$ for all $n$ and both limits exist, then 
$$
\lim x_n\leq\lim y_n.
$$

