Let $x_0$ be an accumulation point of the set $D \subset \mathbb{R}$. We say that $y$ is a limit point of a function $f:D \rightarrow \mathbb{R}$ in $x_0$ iff there exists a sequence $(x_n)$, where $x_n \in D\setminus \{x_0\}$ for $n\in \mathbb{N}$ and $x_n \rightarrow x_0$, such that $y=\lim_{n\rightarrow \infty} f(x_n)$.
I don't know how to prove the following lemma:
If $f: (0,c) \rightarrow \mathbb{R}$ be continuous then the set of all limits points of $f$ in $0$ is the interval $$[\liminf_{x\rightarrow 0}f(x), \limsup_{x\rightarrow 0}f(x)].$$
Best Answer
Note: I assume you require that $x_n\to x_0$ in your definition, otherwise this lemma is false.
Suppose $y$ is a limit point of $f$ in $0$, so we have a sequence $(x_n)$ in $(0,c)$ such that $x_n\to x$ and $f(x_n)\to y$. Then $$\liminf\limits_{n\to\infty} f(x_n)=y=\limsup\limits_{n\to\infty} f(x_n)$$ so we have that $$\liminf\limits_{x\to 0} f(x)\leq \liminf\limits_{n\to\infty} f(x_n)=y=\limsup\limits_{n\to\infty} f(x)\leq \limsup\limits_{x\to 0} f(x)$$ by definition. Thus $y\in [\liminf\limits_{x\to 0} f(x),\limsup\limits_{x\to 0} f(x)]$.
On the other hand, suppose $y\in [\liminf\limits_{x\to 0} f(x),\limsup\limits_{x\to 0} f(x)]$. Take sequence $(x_n)$ and $(x_n')$ in $(0,c)$ converging to $0$ such that $\lim\limits_{n\to\infty} f(x_n)=\liminf\limits_{x\to 0} f(x)$ and $\lim\limits_{n\to\infty} f(x_n')=\limsup\limits_{x\to 0} f(x)$. If $y$ is either $\liminf\limits_{x\to 0} f(x)$ or $\limsup\limits_{x\to 0} f(x)$ we are done. Otherwise, we have some $N$ such that $$n\geq N\implies f(x)<y<f(x_n')$$ so by the Intermediate value theorem we have some $y_n$ between $x_n$ and $x_n'$ such that $f(y_n)=y$. Since $x_n\to 0$ and $x_n'\to 0$ it follows that $y_n\to 0$, and clearly $f(y_n)\to y$, thus $y$ is a limit point of $f$ in $0$.