Answer
(a) The limit exists for all real numbers.
(b) The function is continuous at all real numbers except integers.
The function is discontinuous at integers.
Work Step by Step
(a) $f(x) = \lfloor x \rfloor + \lfloor -x \rfloor$
Suppose $b$ is not an integer.
Let $\lfloor b \rfloor = c$
Then there is a number $\delta \gt 0$ such that $c \lt (b-\delta) \lt b \lt (b+\delta) \lt c+1$
For all numbers $x$ in the interval $(b-\delta, b+\delta)$:
$f(x) = \lfloor x \rfloor + \lfloor -x \rfloor = c+(-c-1) = -1$
Thus $\lim\limits_{x \to b}f(x) = -1,~~~$ and clearly the limit exists.
Suppose $b$ is an integer.
$\lim\limits_{x \to b^-} f(x)$
$= \lim\limits_{x \to b^-}\lfloor x \rfloor + \lfloor -x \rfloor$
$= (b-1)+(-b) = -1$
$\lim\limits_{x \to b^+} f(x)$
$= \lim\limits_{x \to b^+}\lfloor x \rfloor + \lfloor -x \rfloor$
$= (b)+(-b-1) = -1$
Since $\lim\limits_{x \to b}f(x) = -1,~~~$ the limit exists.
The limit exists for all real numbers.
(b) Suppose $b$ is not an integer.
As shown above, $f(b) = -1$
Since $\lim\limits_{x \to b}f(x) = f(b)$, the function is continuous at $b$
Suppose $b$ is an integer.
$f(b) = \lfloor b \rfloor + \lfloor -b \rfloor = b+(-b) = 0$
Since $\lim\limits_{x \to b}f(x) \neq f(b)$, the function is discontinuous at $b$
Therefore, the function is continuous at all real numbers except integers.
The function is discontinuous at integers.
