Show That $\limsup_{x \to x_0} f(x) \ge \liminf_{x \to x_0} f(x)$ My question concerns proving an inequality between two extreme limits, namely:
$$\limsup_{x \to x_0} f(x) \ge \liminf_{x \to x_0} f(x)$$
Using the following defintions:
Let $f: E \to \mathbb{R}$ be a function with domain $E$ in which $x_0$ is a point of accumulation. Then:
$$\limsup_{x \to x_0} f(x) = \inf_{\delta > 0} \sup \left\{f(x): x \in (x_0-\delta, x_0 + \delta) \cap E, \ \ x \neq x_0 \right\}$$
$$\liminf_{x \to x_0} f(x) = \sup_{\delta > 0} \inf \left\{f(x): x \in (x_0-\delta, x_0 + \delta) \cap E, \ \ x \neq x_0 \right\}$$
Using these defintions, show that:
$$\limsup_{x \to x_0} f(x) \ge \liminf_{x \to x_0} f(x)$$ 

It is better to assume that $f$ is bounded to avoid these extreme limits become $\pm\infty$. The proof below can be modified to handle the cases when one or both of these extreme limits are $\pm\infty$.
Note that the function $$A(\delta) = \sup\,\{f(x)\mid x \in (x_{0} - \delta, x_{0} + \delta) \cap E, x \neq x_{0}\}$$ is an increasing function of $\delta$ and and since $f$ is bounded it follows that $\lim_{\delta \to 0^{+}}A(\delta)$ exists and is equal to $\inf_{\delta > 0}A(\delta)$. Similarly the function $$B(\delta) = \inf\,\{f(x)\mid x \in (x_{0} - \delta, x_{0} + \delta) \cap E, x \neq x_{0}\}$$ is a decreasing function of $\delta$ and the limit $\lim_{\delta \to 0^{+}}B(\delta)$ exists and is equal to $\sup_{\delta > 0}B(\delta)$.
Thus following your definitions (and reasoning in previous paragraph) we have $$\limsup_{x \to x_{0}}f(x) = \lim_{\delta \to 0^{+}}A(\delta),\, \liminf_{x \to x_{0}}f(x) = \lim_{\delta \to 0^{+}}B(\delta)$$ Since we have $A(\delta) \geq B(\delta)$ for all values of $\delta > 0$, it follows that their limits also follow the same inequality and hence $\limsup f(x) \geq \liminf f(x)$.
