Definition: The limit inferior of a function $f(x)$ is defined as:
$$\underline{\lim}_\limits{x\rightarrow 0}f(x)=\inf\{m\,:\,\exists\ x_n\rightarrow 0\text{ with }f(x_n)\rightarrow m\}.$$
Show that if $f(x)\leq g(x)$ Then $\underline{\lim}_{x\rightarrow 0}f(x)\leq\underline{\lim}_{x\rightarrow 0}g(x)$ (Assuming limits exist)
May I have hints on how to prove this, please?
The idea I have is to start with a contradiction but then I am not sure what to do. The idea seems reasonably clear for continuous function, and geometrically this is clear, however I would really like to know how to start a formal proof.
Suppose that $\underline{\lim_{x\to0}}f(x)>\underline{\lim_{x\to0}}g(x)$. Take$$m\in\left(\underline{\lim_{x\to0}}g(x),\underline{\lim_{x\to0}}f(x)\right).$$It follows from the definition of $\underline{\lim_{x\to0}}g(x)$ and from the fact that $m>\underline{\lim_{x\to0}}g(x)$ that there is a sequence $(x_n)_{n\in\mathbb N}$ such that $\lim_{n\to\infty}x_n=0$ and that the sequence $\bigl(g(x_n)\bigr)_{n\in\mathbb N}$ converges to some $m^\ast\leqslant m$. The sequence $\bigl(f(x_n)\bigr)_{n\in\mathbb N}$ is bounded and therefore there is a subsequence $(x_{n_k})_{k\in\mathbb N}$ of $(x_n)_{n\in\mathbb N}$ such that $\bigl(f(x_{n_k})\bigr)_{k\in\mathbb N}$ converges. But its limit is smaller than or equal to $m^\ast$ (since $f(x_{n_k})\leqslant g(x_{n_k})\rightarrow m^\ast$). This is impossible, since$$m^\ast\leqslant m<\underline{\lim_{x\to0}}f(x).$$