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

Question

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.

Answer 1

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 $\lim_{n\to\infty}f\left(\frac1n\right)=0$.

Answer 2

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 $x_0=0$.