Answer
$\lim\limits_{x \to 2}f(x) = -1$
$f(2) = 0$
$\lim\limits_{x \to 2}f(x) \neq f(2)$
Work Step by Step
$f(x) = \lfloor{x}\rfloor + \lfloor{-x}\rfloor$
We can find $f(2)$:
$f(2) = \lfloor{2}\rfloor + \lfloor{-2}\rfloor$
$f(2) = 2 + (-2)$
$f(2) = 0$
We can find $\lim\limits_{x \to 2^-}f(x)$:
$\lim\limits_{x \to 2^-}(\lfloor{x}\rfloor + \lfloor{-x}\rfloor)$
$=\lim\limits_{x \to 2^-}\lfloor{x}\rfloor +\lim\limits_{x \to 2^-}\lfloor{-x}\rfloor$
$=\lim\limits_{x \to 2^-}1 +\lim\limits_{x \to 2^-}-2$
$= 1+(-2)$
$= -1$
We can find $\lim\limits_{x \to 2^+}f(x)$:
$\lim\limits_{x \to 2^+}(\lfloor{x}\rfloor + \lfloor{-x}\rfloor)$
$=\lim\limits_{x \to 2^+}\lfloor{x}\rfloor +\lim\limits_{x \to 2^+}\lfloor{-x}\rfloor$
$=\lim\limits_{x \to 2^+}2 +\lim\limits_{x \to 2^+}-3$
$= 2+(-3)$
$= -1$
Since $\lim\limits_{x \to 2^-}f(x) = \lim\limits_{x \to 2^+}f(x)$, then $\lim\limits_{x \to 2}f(x)$ exists.
However, $\lim\limits_{x \to 2}f(x) = -1$ but $f(2) = 0$
That is, $\lim\limits_{x \to 2}f(x) \neq f(2)$
