Answer
$a.\quad $yes
$b.\quad $yes
$c.\quad $no
$d.\quad $no
Work Step by Step
The graph of f(x) contains points $(x,f(x))$, where x belongs to the domain of f.
$a.$
The point $(1,1)$ belongs to the graph of f, so $f(1)=1$
( $f(1)$ exists)
$b.$
As x approaches the value 1 from the left, f(x) approaches 2, and, as x approaches the value 1 from the right, f(x) approaches 2.
The one-sided limits exist, and are equal,
$\displaystyle \lim_{x\rightarrow 1^{-}}f(x)=\lim_{x\rightarrow 1^{+}}f(x)=2$,
so, by Th.6 (sec 2.4), a limit at x=1 exists (and equals 2),
$\displaystyle \lim_{x\rightarrow 1}f(x)=2$
$c.$
$\displaystyle \lim_{x\rightarrow 1}f(x)=2\ \ \neq\ \ f(1)=1$
(they are not equal)
$d.$
Apply "Continuity test", pg.95
1. $f(c)$ exists ( $c$ lies in the domain of $f$).
2. $\displaystyle \lim_{x\rightarrow c}f(x)$ exists ( $f$ has a limit as $x\rightarrow c$).
The first two conditions are satisfied at $x=1$
BUT the third one is not:
3. $\displaystyle \lim_{x\rightarrow c}f(x)=f(c)$ (the limit equals the function value).
Therefore, f is not continuous at $x=1$.