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!


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

  • $\begingroup$ 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

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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