Let $f: X \to \mathbb R$. Define $$\liminf_{x \to x_0} f(x) = \lim_{r \to 0^+} \inf_{x \in B_r(x_0)} f(x)$$ Show if $x_n \to x_0$ then $$\liminf_{n \to \infty} f(x_n) \ge \liminf_{x \to x_0} f(x)$$ and show there exists a sequence $x_n$ converging to $x_0$ such that $$\lim_{n \to \infty} f(x_n) = \liminf_{x \to x_0} f(x)$$
Since $f$ is arbitrary we don't have a nice result like $f(x_n) \to f(x_0)$ which would make the problem easy. I believe if for all $n \ge N, d(x_n, x_0) < \epsilon$ then $\inf_{n \ge N} f(x_n) \ge \inf_{x \in B(x_0, \epsilon)} f(x)$ just because the ball encompasses more points. However I am not sure about the limiting behavior of both sides, that is if it holds for all $\epsilon$ then does it hold for the limit? Also, can we pick our $x_n$ such that $\inf f(x_n) = \inf_{x \in B(x_0, \epsilon)} f(x)$? Any hints or pointers are appreciated.