Answer
(a) (i) $\lim\limits_{x \to -2^+}\lfloor{x}\rfloor = -2$
(ii) $\lim\limits_{x \to -2}\lfloor{x}\rfloor $ does not exist.
(iii)$\lim\limits_{x \to -2.4}\lfloor{x}\rfloor = -3$
(b) (i) $\lim\limits_{x \to n^-}\lfloor{x}\rfloor = n-1$
(ii) $\lim\limits_{x \to n^+}\lfloor{x}\rfloor = n$
(c) $\lim\limits_{x \to a}\lfloor{x}\rfloor$ exists if $a$ is not an integer.
If $a$ is an integer, then the limit does not exist.
Work Step by Step
(a) (i) $\lim\limits_{x \to -2^+}\lfloor{x}\rfloor = \lim\limits_{x \to -2^+} -2 = -2$
(ii) $\lim\limits_{x \to -2^-}\lfloor{x}\rfloor = \lim\limits_{x \to -2^-} -3 = -3$
Since $\lim\limits_{x \to -2^-}\lfloor{x}\rfloor \neq \lim\limits_{x \to -2^+}\lfloor{x}\rfloor$, then $\lim\limits_{x \to -2}\lfloor{x}\rfloor $ does not exist.
(iii)$\lim\limits_{x \to -2.4}\lfloor{x}\rfloor = \lim\limits_{x \to -2.4} -3 = -3$
(b) Suppose that $n$ is an integer.
(i) $\lim\limits_{x \to n^-}\lfloor{x}\rfloor = \lim\limits_{x \to n^-} (n-1) = n-1$
(ii) $\lim\limits_{x \to n^+}\lfloor{x}\rfloor = \lim\limits_{x \to n^+} n = n$
(c) If $a$ is an integer, then $\lim\limits_{x \to a^-}\lfloor{x}\rfloor \neq \lim\limits_{x \to a^+}\lfloor{x}\rfloor$.
Therefore, $\lim\limits_{x \to a}\lfloor{x}\rfloor$ does not exist if $a$ is an integer.
If $a$ is not an integer, then $\lim\limits_{x \to a^-}\lfloor{x}\rfloor = \lim\limits_{x \to a^+}\lfloor{x}\rfloor = \lfloor{a}\rfloor$
Therefore, $\lim\limits_{x \to a}\lfloor{x}\rfloor$ exists if $a$ is not an integer.
