Answer
$a)$
i) $\lim\limits_{x \to -2^-} f(x)= -1$
ii)$\lim\limits_{x \to -2^+} f(x)= \frac{-1}{2}$
iii)$\lim\limits_{x \to -2} f(x)$ does not exist.
iv) $f(a)$ does not exist.
$b)$
i) $\lim\limits_{x \to -1^-} f(x)= \frac{-1}{2}$
ii)$\lim\limits_{x \to -1^+} f(x)= \frac{-1}{2}$
iii)$\lim\limits_{x \to -1^-} f(x)= \frac{-1}{2}$
iv) $f(-1) = \frac{-1}{2}$
Work Step by Step
$a)$
i) As we approach the $x$-value of $-2$ from the L.H.S, $f(x)$ approaches $-1$.
$\displaystyle\lim_{x\rightarrow -2^-}f(x)=-1$
ii) As we approach the $x$-value of $-2$ from the R.H.S, $f(x)$ approaches $\frac{-1}{2}$
$\displaystyle\lim_{x\rightarrow -2^+}f(x)=-\frac{1}{2}$
iii) Since the left-hand limit and right-hand limit of the function are not equal, $\lim\limits_{x \to -2}f(x)$ does not exist.
iv) The function is undefined at $x = -2$.
$b)$
i) As we approach the $x$-value of $-1$ from the L.H.S, $f(x)$ approaches $\frac{-1}{2}$.
$\displaystyle\lim_{x\rightarrow -1^-}f(x)=-\frac{1}{2}$
ii) As we approach the $x$-value of $-1$ from the R.H.S, $f(x)$ approaches $\frac{-1}{2}$.
$\displaystyle\lim_{x\rightarrow -1^+}f(x)=-\frac{1}{2}$
iii) Since the lift-hand limit and right-hand limit of the function are equal at $x = -1$, $\lim\limits_{x \to -1} f(x) = \frac{-1}{2}$.
iv) The function has the value of $\frac{-1}{2}$ at $x = -1$.
