Prove $\lim \min(a, a_n) = a$ if $a \leq \lim \inf a_n$

Question

I'm trying to prove that: if $a \leq \lim\limits_{n \to \infty} \inf a_n$ then $$\lim\limits_{n \to \infty} \min (a, a_n) = a$$ It's direct to see that $\lim\limits_{n \to \infty} \min (a, a_n) \leq a$, but I still struggle to prove the other direction.

Answer 1

Let $\theta=\liminf a_n<\infty$ and let $\varepsilon>0$. Then there exists $n_0\in\mathbb{N}$ such that for all $n\geq n_0$ we have that $a_n\geq\theta-\varepsilon$. Hence, for $n\geq n_0: \min(a,a_n)\geq\min(a,\theta-\varepsilon)\geq\min(a,a-\varepsilon)=a-\varepsilon$. The case $\liminf=\infty$ is easier since then $a_n\geq a$ for large enough $n$.