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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):

  1. 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 $0 \lt \infty$.

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

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

  1. 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!

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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 $$ \lim x_n\leq\lim y_n. $$

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  • $\begingroup$ I see! thank you! may I ask you a final question? then in my question math.stackexchange.com/q/1937867/189215 I am starting with $\exists p \in \Bbb P: n^3 \lt p \lt (n+1)^3$ so according to your explanation, if I start instead with $\exists p \in \Bbb P: n^3 \le p \le (n+1)^3$ which is also true, then I can assure that indeed as I was expecting $\lim_{n \to \infty} log_np=3$. Is that right? $\endgroup$ – iadvd Sep 23 '16 at 2:32
  • $\begingroup$ Yes. Of course, $p$ depends on $n$, so it's not like you are getting much information: if a number is close to $n^3$, then its $\log_n$ will be close to $3$. $\endgroup$ – Martin Argerami Sep 23 '16 at 3:27
  • $\begingroup$ thanks for the confirmation! Yes is quite trivial, just reformulating the initial inequality and observing what happens at the limit! but the same can not be said for $n^2$ intervals because it is not know if Legendre's conjecture is true. So we do not now if $lim_{n\to\infty} log_n p = 2$ because it is not known if always exists $p$ in $[n^2,(n+1)^2]$. I was just trying to see those problems with a different perspective. $\endgroup$ – iadvd Sep 23 '16 at 3:33
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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.

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