If $f(x)$ is continuous on interval $(0, +\infty)$ and $\lim_{x\to +\infty}{f(x)} \in \mathbb {R}$ exists, is it true that $\lim_{n \to \infty}{(x_n)}$ also exists for $(x_n)=f(nx_0)$ and any $x_0>0$? I realized that if $f(x)$ is not continuous then the converse is not true (for example, we can consider $f(x)=\frac{x}{2}$ if $x$ is rational and $f(x)=x$ otherwise). Also I realized that the converse is always true for continuous $f(x)$.
But I can't realize is my initial supposition stated in this question is true.
Let $\epsilon >0$ and $d=\lim_{x \to \infty } f(x)$. There exists $A \in (0,\infty )$ such that $x >A$ implies $|f(x)-d| <\epsilon $. If $ n>\frac A {x_0}$ then $|f(nx_0)-d| <\epsilon $ because $nx_0 >A$ Taking $n_0$ to be least integer greater than $\frac A {x_0}$ we get $|f(nx_0)-d| <\epsilon $ for all $n \geq n_0$. Hence $\lim_{n \to \infty} f(nx_0) =d$. Continuity of $f$ is not required for this.