I know that, given a function $f:\mathbb{R} \to\mathbb{R}$ we have that: $$ \lim _{x \to x_0} f(x)=\lambda \iff \lim _{n \to +\infty} f(x_n)=\lambda $$ for every sequence $(x_n)_{n \in \mathbb{N}}$ such that $x_n \to x_0$.
I was wondering if we can say the same thing for $\liminf$ and $\limsup$, considering the two different definitions: $$\limsup_{n \to +\infty} f(x_n)=\inf_{n \in \mathbb{N}} \left( \sup _{k \ge n} f(x_k)\right), \quad \limsup _{x \to x_0} f(x)=\lim _{r \to 0^+} \left( \sup _{x \in B(x_0,r)} f(x) \right)$$ And analogous definition for the $\liminf$. In other words, I was wondering if it's true that: $$\limsup _{x \to x_0} f(x)=\lambda \iff \limsup _{n \to +\infty} f(x_n)=\lambda$$ for every sequence $(x_n)_{n \in \mathbb{N}}$ such that $x_n \to x_0$ And same for the $\liminf$.
Edit: because of the answer by Goonfiend, which provided an example of $$\liminf _{x \rightarrow x_0} f(x) < \liminf _{n \rightarrow +\infty} f(x_n)$$ I tried to prove that $$\liminf _{x \rightarrow x_0} f(x) \le \liminf _{n \rightarrow +\infty} f(x_n) \ , \ \limsup _{x \rightarrow x_0} f(x) \ge \limsup _{n \rightarrow +\infty} f(x_n) $$ for every sequence $x_n \rightarrow x_0$.
Proof: because of the above characterization of limits, we have that $$\quad \limsup _{x \to x_0} f(x)=\lim _{r \to 0^+} \left( \sup _{x \in B(x_0,r)} f(x) \right)=\lim _{n\rightarrow +\infty}\left( \sup _{x \in B(x_0,r_n)} f(x) \right)= \inf _{n \in \mathbb{N}} \left( \sup _{x \in B(x_0,r_n)} f(x) \right)$$ Where $(r_n)_{n \in \mathbb{N}}$ is an arbitrary decreasing infinitesimal sequence ($r_n \searrow 0$). Then we can choose $r_n$ such that $x_k \in B(x_0,r_n)$ for every $k \ge n$. This means that $\{x_k \mid k \ge n\} \subseteq B(x_0,r_n)$, and so we have that $$\sup _{k \ge n} f(x_k) \le \sup _{x \in B(x_0,r_n)} f(x) \quad \forall n \in \mathbb{N} \Rightarrow \inf_{n \in \mathbb{N}} \left( \sup _{k \ge n} f(x_k)\right) \le \inf_{n \in \mathbb{N}} \left( \sup _{x \in B(x_0,r_n)} f(x)\right)$$ And this means, by definition, that $$\limsup _{x \rightarrow x_0} f(x) \ge \limsup _{n \rightarrow +\infty} f(x_n)$$
Is this a correct proof?
No it is not true. Consider $f(x)=1_\mathbb{Q}$. Then $$\liminf_{x\to 0}f(x)=0\neq 1= \liminf_{n\to \infty} f(1/n).$$ It is false for lim sup too for similar reasons.