Answer
(a) Below, we can see a graph below of $f(x) = \lfloor cos~x \rfloor$ on the interval $-\pi \leq x \leq \pi$
(b) (i) $\lim\limits_{x \to 0}f(x) = 0$
(ii) $\lim\limits_{x \to (\pi/2)^-}f(x) = 0$
(iii) $\lim\limits_{x \to (\pi/2)^+}f(x) = -1$
(iv) $\lim\limits_{x \to \pi/2}f(x)$ does not exist
(c) $\lim\limits_{x \to a}f(x)$ exists for all values of $a$ in the interval except $-\frac{\pi}{2}$ and $\frac{\pi}{2}$
Work Step by Step
(a) Below, we can see a graph of $f(x) = \lfloor cos~x \rfloor$ on the interval $-\pi \leq x \leq \pi$
(b) We can evaluate each limit:
(i) $\lim\limits_{x \to 0}f(x) = 0$
(ii) $\lim\limits_{x \to (\pi/2)^-}f(x) = 0$
(iii) $\lim\limits_{x \to (\pi/2)^+}f(x) = -1$
(iv) $\lim\limits_{x \to \pi/2}f(x)$ does not exist because $\lim\limits_{x \to (\pi/2)^-} f(x) \neq \lim\limits_{x \to (\pi/2)^+} f(x)$
(c) $\lim\limits_{x \to a}f(x)$ exists for all values of $a$ in the interval $-\pi \leq x \leq \pi$ except where the left limit is not equal to the right limit. Therefore, the limit exists for all values of $a$ except $-\frac{\pi}{2}$ and $\frac{\pi}{2}$
