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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.

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  • $\begingroup$ The answer to the first question is yes and it follows by definition of limit. $\endgroup$
    – Kavi Rama Murthy
    Jun 25, 2018 at 9:54
  • $\begingroup$ @KaviRamaMurthy I also think so, but I can't prove it elegantly, using continuity of $f(x)$. $\endgroup$
    – Tyrone Ward
    Jun 25, 2018 at 10:02
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    $\begingroup$ 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 $f(x_n) \to A$. $\endgroup$
    – Botond
    Jun 25, 2018 at 10:03

1 Answer 1

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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.

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