$f$ be a function defined on some subset $D$. The function has a limit $l$ at $0$ iff for every sequence $(x_n)$ in $D$ converging to $0$, the sequence $f(x_n)$ converges to $l$.

Now, if for any sequence of rational numbers $(a_n)$ and any sequence of irrational numbers $(b_n)$ converging to $0$, if $f(a_n)$ and $f(b_n)$ converge to $l$, does that imply the limit at $0$ is $l$? I know that I left out those sequences which have both rational and irrational elements. Just wondeing if the limit exists in such a case. Thanks in advance!


2 Answers 2


Yes. We generally name a sequence that converges to zero a null sequence.

Observe that any null sequence can be decomposed in two ordered lists with only rational and irrational numbers, where at least one of them is infinite, that is, at least one of them is a subsequence of the original sequence.

  • If one of the ordered list is finite then your hypothesis holds, because at some point your null sequence will had only rational or irrational numbers.

  • If the two ordered lists are infinite then your hypothesis also holds: let $(a_n)$ a null sequence such that it can be decomposed in two null subsequences $(a_{g(n)})$ and $(a_{h(n)})$ were the first are only rational numbers and the second are only irrational numbers (where $g,h:\Bbb N\to\Bbb N$ are increasing functions that defines it indices).

    By assumption we knows that $\lim f(a_{g(n)})=\lim f(a_{h(n)})=l$ and we need to show that $\lim f(a_n)=l$. Then for any chosen $\epsilon>0$ exists $N_1,N_2\in\Bbb N$ such that

    $$|f(a_{g(n)})-l|<\epsilon,\quad\forall n\ge N_1\\|f(a_{h(n)})-l|<\epsilon,\quad\forall n\ge N_2$$

    Then choosing $N:=\max\{g(N_1),h(N_2)\}$ we have that

    $$|f(a_n)-l|<\epsilon,\quad\forall n\ge N$$

  • $\begingroup$ Is this only true for null sequences though? My thought process went like this : let $(x_n)$ be a a sequence converging to 0. $(f(x_n))$ is certainly not properly divergent. If possible let it have more than one limit points. Then corresponding to that there is a subsequence converging to that point and thus every subsequence of it consisting solely of rationals or irrationals will also converge there; which means the only limit point is $l$ $\endgroup$
    – Not Euler
    Jan 31, 2018 at 21:51
  • $\begingroup$ a sequence that converges to zero is a null sequence. Your question is about null sequences, but suppose that all assumptions are equal to your asked question except that you are talking of sequences that converge for some other arbitrary (but fixed) value different from zero. Then all the same explained in my answer holds. $\endgroup$
    – Masacroso
    Jan 31, 2018 at 21:57
  • $\begingroup$ thanks for the answer :) $\endgroup$
    – Not Euler
    Jan 31, 2018 at 21:59

Let $x_n\to 0$. Further let $a_{n_k}$ be the rational subsequence and $b_{p_k}$ be the irrational subsequence of $x_n$. If $a_{n_k}$ or $b_{p_k}$ is a finite sequence, then there is nothing to show. Assume they are infinite.

Let $\epsilon > 0$. Then, there exists some $k_0$ with $$ |f(a_{n_k}) - l| < \epsilon \text{ and } |f(b_{p_k}) - l| < \epsilon $$ for every $k\ge k_0$. Can you give some $n_0$ such that $$ |f(x_n) - l| < \epsilon $$ holds for every $n\ge n_0$?


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