We start with a definition.
We say that a function $f\colon \mathbb{R}^n\to \mathbb{R}$ is lower semi continuous (l.s.c.) if $$\forall x\in\mathbb{R}^n,\; \liminf_{y\to x}f(y)\geq f(x).$$
As homework, I was assigned to prove that
$f$ is l.s.c. iff $\; \forall x_n\to x,\; \lim_{n\to\infty}f(x_n)\geq f(x)$ $\qquad(*)$.
yet I came up with the following counter example which shows that there exists a $f\colon\mathbb{R}^n\to\mathbb{R}$ l.s.c. and $x_n\to x$ such that there is no $l$ with $f(x_n)\to l$:
Let $f\colon \mathbb{R}\to \mathbb{R}$ be defined by $f(x)=\begin{cases}1, & x\neq0\\ 0, & x=0\end{cases}$. Then $\forall x\in\mathbb{R},\; \liminf_{y\to x}f(y)=1\geq f(x)$ and $f$ is l.s.c. Consider $x_n=\frac{1+(-1)^n}{2}\frac{1}{n}$ so that $(x_n)_{n\in\mathbb{N}}=(0,\frac{1}{2},0,\frac{1}{4},0,\frac{1}{6},0,\dots)$. Clearly $0\leq x_n\leq \frac{1}{n}$ so $x_n\to 0$ yet $\big(f(x_n)\big)_{n\in\mathbb{N}}=(0,1,0,1,0,1,\dots)$ wich doesn't converge so $\lim_{n}f(x_n)\geq f(0)=0$ can't be true and $(*)$ is false.
Am I being asked to prove something that is false? Is the counter example correct? On the statement of the property, is it implied that $f(x_n)$ converges? (that is, the statement is meant to be interpreted as $\forall x\to x$ such that $f(x_n)$ converge, $\lim f(x_n)\geq f(x)$. In which case it seems to be true). Thanks in advance.