A counterexample of the converse of the theorem of sequence and its function

Question

I trying to find a counterexample of the converse of the following theorem:

Given a sequence $a_{n}$ if we have a function $f(n)=a_{n}$ and $\lim_{x\to\infty} f(x)=a$, then $\lim_{n\to\infty} an=a$.

That is I want to find a counterexample of the following statement:

Given a sequence $a_{n}$ and $\lim_{n\to\infty} a_n=a$,if we have a function $f(n)=a_{n}$ then $\lim_{x\to\infty} f(x)=a$

I think $a_n=|(-1)^n|$ is one of those counterexamples since the limit of this sequence is 1, but I am not sure is $\lim_{x\to\infty}|(-1)^x|$ exists or not. Here is my approach: $$ \lim_{x\to\infty}|(-1)^x|=\lim_{x\to\infty} |cos(\pi x)+i\cdot sin(\pi x)| $$ Since the limit of cosine and sine are divergent, can we say this limit is divergent? Can anyone explain to me and point out where I go wrong(P.S. I have not learn Complex Analysis yet)? Or can anyone provide another counterexample of this statement? Thank you very much.

Answer 1

The point is very simple : $a_n$ is some sequence which determines $f$ only at the positive integers, and $f$ is fairly arbitrary except at the positive integers. It is not even given to be continuous : this allows you to play to the weaknesses of $f$ cleverly.

You are given $a_n \to a$, and the converse statement is that for all $f$ such that $f(n) = a_n$,we have existence of the limit $\lim_{x \to \infty} f(x)$. It's mathematical negation, which is to be proved, is : there is some covergent sequence $a_n \to a$ and some function $f$ such that $f(n) = a_n$ but $\lim_{x \to \infty} f(x)$ does not exist.

We will show a stronger statement : that for all (not just for one, like in the negation) $a_n \to a$, there is $f$(depending on $a_n$) such that $f(n) = a_n$ but $\lim_{x \to \infty} f(x)$ does not exist.

It is quite clear how we do this. Given $a_n$ with limit $a$, define $f : \mathbb{R} \to \mathbb{R}$ in the following way : fix $b \neq a$. Define $f(n) = a_{|n|}$ for $n$ any nonzero integer, $f(0) = 0$ and $f(x) = b$ for $x$ non-integral.

Now, of course $f(n) = a_n$ for all positive integers $n$ ,but $\lim_{x \to \infty} f(x)$ does not exist, since if $\lim_{x \to \infty} f(x) = c$, then fix $\epsilon > 0$. By the definition, there exists $r$ such that if $x > r$ then $|f(x) - c| < \epsilon$. Call this statement $(*)$.

Find an positive integer $N$ greater than $r$, then this says $|f(n) - c| < \epsilon$ for all $n > N$, which we rephrase : for all $\epsilon > 0$, there is $N \in n$, such that $n > N \implies |a_n - c| < \epsilon$. This matches with the definition of limit ! So, $a_n \to c$. But by uniqueness of limits, $c = a$.

By $(*)$, find some non-integer $m > r$. Then, $|f(m) - c| = |b - c| = |b-a|< \epsilon$ for all $\epsilon > 0$, which leads to $|b-a| = 0 \implies b = a$, a contradiction.

Hence, the statement is false, with the given strong family of counterexamples.