Does the chain of inequalities $\liminf_{n\to +\infty} f(x_n)\le f(x_n)\le\limsup_{n\to +\infty} f(x_n)$ hold true?

Question

Let $f$ be a real valued function. Let $(x_n)_n$ be a real sequence.

Is true that $$\liminf_{n\to +\infty} f(x_n)\le f(x_n)\le\limsup_{n\to +\infty} f(x_n)?$$

I know that for every sequence $$\liminf_{n\to +\infty} x_n\le\limsup_{n\to +\infty} x_n,$$ but I don't know how to prove that the former chain of inequalities is true.

Could someone please help?

Thank you in advance.

Answer 1

No, the “chain of inequalities’’ is false. It can even be false for infinitely many $n$; consider the sequence $x_n=(-1)^n/n$.

Answer 2

No. Consider the sequence $(x_n)$, where $x_n=(-1)^n2^{-n}$, and let $f$ be the function $f(x)=x$. Check to see that the proposed string of inequalities does not hold for any nonnegative integer $n$.

This extends to contiuous functions. Graph, say $e^{-x}\sin(x)$, or even $x^2e^{-x}\sin(x)$.