• f(x) = 2 if x=3; x−1 if x>3; (x+3)/3 if x<3
• Bf(x) = 4 if x=3; 8−x if x≠3
• Cf(x) = x+3 if x≤3; x−4 if x>3
• Df(x) = 1/(x³−27) if x≠3

For option A: f(3)=2, the right-hand limit as x→3⁺ of (x−1) is 2, and the left-hand limit as x→3⁻ of (x+3)/3 is 6/3=2 — all three agree, so it's continuous at x=3. For B, f(3)=4 but the limit of 8−x as x→3 is 5≠4. For C, f(3)=6 (from x+3) but the right-hand limit of x−4 is −1≠6. For D, the function isn't even defined at x=3 (denominator x³−27=0 there) and blows up nearby, so it's discontinuous. Only A is continuous at x=3.