# Is it true that if continuous $f(x)$ has a limit then $f(nx_0)$ has a limit for any $x_0$?

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.

• The answer to the first question is yes and it follows by definition of limit. Jun 25, 2018 at 9:54
• @KaviRamaMurthy I also think so, but I can't prove it elegantly, using continuity of $f(x)$. Jun 25, 2018 at 10:02
• Yes. Let $f: \mathbb{R} \to \mathbb{R}$ be a function and $a \in (\text{dom}(f))'$. Then the following statements are equivalent: (1) $\exists \lim\limits_{a}f=:A$ (2) For every $(x_n)_{n \in \mathbb{N}}$ sequence in $\text{dom}(f)$, with $x_n \to a$, and $\forall n \in \mathbb{N}, x_n \neq a$, we have that $f(x_n) \to A$. Jun 25, 2018 at 10:03

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.