Answer
(a)$$D_f= (- \infty , 0] \cup [1, + \infty )$$
(b)$$\lim_{x \to 0^-}\sqrt{x(x-1)}=0$$
(c)$$\lim_{x \to 1^+}\sqrt{x(x-1)}=0$$
Work Step by Step
(a)
The domain of a square root function is the set of real numbers making the radicand non-negative. Thus, we have$$x(x-1) \ge 0 \quad \Rightarrow \quad \begin{cases}x \ge 0, \\ x-1 \ge 0 \end{cases} \quad \text{or} \quad \begin{cases} x \le 0, \\ x-1 \le 0 \end{cases} \quad \Rightarrow \quad \begin{cases}x \ge 0, \\ x \ge 1 \end{cases} \quad \text{or} \quad \begin{cases} x \le 0, \\ x \le 1 \end{cases} \quad \Rightarrow \quad D_f= (- \infty , 0] \cup [1, + \infty ).$$
(b)
$$\lim_{x \to 0^-}\sqrt{x(x-1)}=\sqrt{0(0-1)}=0$$
(c)
$$\lim_{x \to 1^+}\sqrt{x(x-1)}=\sqrt{1(1-1)}=0$$
