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Let $f:\mathbb{R} \to \mathbb{R}$. Can we say that $\lim\limits_{n \to \infty}f(x_0+\frac{1}{n})=l$ is another way of expressing the right-sided limit at $x_0$?
I tried to use the definition of the limit,but I am stuck.Intuitively, it seems true, but I don't know how to prove it.

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No, it is not true. Take, for instance,$$f(x)=\begin{cases}\sin\left(\frac\pi x\right)&\text{ if }x\neq0\\0&\text{ if }x=0.\end{cases}$$Then the limit $\lim_{x\to0^+}f(x)$ doesn't exist, in spite of the fact that $\lim_{n\to\infty}f\left(\frac1n\right)=0$.

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    $\begingroup$ Who said that $n$ was an integer ? ;-) $\endgroup$ – Yves Daoust Apr 19 at 10:35
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    $\begingroup$ LOL. This is why I prefer the notation for limits that we use in my country: the limit of a sequence $(a_n)_{n\in\mathbb N}$ is denoted by $\lim_{n\in\mathbb N}a_n$, instead of $\lim_{n\to\infty}a_n$. So, there is no ambiguity. $\endgroup$ – José Carlos Santos Apr 19 at 10:39
  • $\begingroup$ @YvesDaoust it is an integer. In my country, integers are commonly denoted by $n$, this is why I forgot that this may not be the case everywhere. $\endgroup$ – Math Guy Apr 19 at 11:42
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    $\begingroup$ @MathGuy: in my country, "n" stands for "not-necessarily-natural" and is short for "nnn" :-) $\endgroup$ – Yves Daoust Apr 19 at 11:47
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No. Right hand limit of $f$ at $x_0$ is $l$ if $f(x_0+r_n) \to l$ for every sequence $(r_n)$ decreasing to $0$. A particular sequence will not do. For example, take $f(x)=0$ for $x$ rational and $1$ for $x$ irrational. Take $x_0=0$.

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