My textbook says $$\limsup_{x \to x_0} f(x) = \max\{f(x_0), \lim_{h\to 0^+} f( x_0 + h), \lim_{h\to 0^-} f( x_0 + h)\}$$
Assuming $f(x_0)$ is distinct from the latter two values, how can $\limsup f(x)$ as $x$ approaches $x_0$ equal neither $\limsup f(x)$ approaching $x_0$ from the left or from the right? Doesn't that violate the definition of a limit, since $f(x_0)$ isn't "approached?"
For clarifcation, assuming wlog $f(x_0) < f(x)$, $\sup f(x) = \max\{(f(x_0), f(x))\}$. The limit in question is the value approached as $x$ approaches $x_0$.
Answer: $\limsup f(x)$ has to include $f(x_0)$ in order for the statement "if $\limsup f(x_0) = \liminf f(x_0)$, then $f$ is continuous at $x_0$" to be always true.
Try $$f(x) = \begin{cases} \sin(1/x) &, x \neq 0 \\ 2 &, x = 0 \end{cases} \text{.} $$ What should $\limsup_{x \rightarrow 0} f(x)$ be?
Or to put this another way, why do you think $x_0$ cannot be a point in the sequence implicitly encoded by $x \rightarrow x_0$? We have to go to some pains to dodge this point in the one-sided limits. (We don't take sequences of points, we take sequences of offsets, $h$ all required to be of the same sign, hence nonzero.)