Answer
$a)$
i) $\lim\limits_{x \to 1^-} = 1$
ii) $\lim\limits_{x \to 1^+} = 1$
iii) $\lim\limits_{x \to 1} = 1$
iv) $f(1) = 2$
$b)$
i) $\lim\limits_{x \to 2^-} = 0$
ii) $\lim\limits_{x \to 2^+} = 0$
iii) $\lim\limits_{x \to 2} = 0$
iv) $f(2) = 0$
Work Step by Step
$a)$
i) As we approach the $x$-value of $1$ from the L.H.S, $f(x)$ approaches $1$.
$\displaystyle\lim_{x\rightarrow 1^-}f(x)=1$
ii) As we approach the $x$-value of $1$ from the R.H.S, $f(x)$ approaches $1$.
$\displaystyle\lim_{x\rightarrow 1^+}f(x)=1$
iii) Since left-hand limit = right-hand limit, the limits exists at $x=1$.
iv) Even though the value of $f(x)$ achieved while approaching $1$ is $1$, $f(x)$ is itself defined at $(1,2)$ and not $(1,1)$ (input $1$, function outputs $2$.)
$b)$
i) As we approach the $x$-value of $2$ from the L.H.S, $f(x)$ approaches $0$.
$\displaystyle\lim_{x\rightarrow 2^-}f(x)=0$
ii) As we approach the $x$-value of $2$ from the R.H.S, $f(x)$ approaches $0$.
$\displaystyle\lim_{x\rightarrow 2^+}f(x)=0$
iii) Since left-hand limit = right-hand limit, the limits exists at $x=2$.
iv) At $x = 2$, the function $f(x)$ is defined (input $2$, function outputs $0$.)
