#### Answer

a. Domain $[0,2]$, Range $y\in(1,1]$ and $y=2$
b. $(0,1)\cup(1,2)$
c. $x=2$
d. $x=0$

#### Work Step by Step

We are given the piecewise function:
$f(x)=\begin{cases} \sqrt {1-x^2}\hspace0.8cm 0\leq x\lt1 \\ 1\hspace2cm 1\leq x\lt2 \\2\hspace2cm x=2 \end{cases}$,
We can graph the function as shown.
a. Domain $[0,2]$, Range $y\in(1,1]$ and $y=2$
b. Check the end points; we see that limits do not exist at $x=0,1,2$, thus $\lim_{x\to c}f(x)$ exist for points in $(0,1)\cup(1,2)$
c. For the left-hand limit only, we see a point at $x=2$ with $\lim_{x\to 2^-}f(x)=1$
d. For the right-hand limit only, we see a point at $x=0$ with $\lim_{x\to 0^+}f(x)=1$