# Is “$\lim\limits_{n \to \infty}f(x_0+\frac{1}{n})=l$” another way of expressing the right-sided limit?

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.

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 $$\lim_{n\to\infty}f\left(\frac1n\right)=0$$.
• Who said that $n$ was an integer ? ;-) – Yves Daoust Apr 19 at 10:35
• 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. – José Carlos Santos Apr 19 at 10:39
• @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. – Math Guy Apr 19 at 11:42
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 $$x_0=0$$.