For continuity, Left hand limit must be equal to right hand limit. For continuity at $x = 3$,
the value of $f(x)$ just above and just below $3$ must be the same.
A. $f(3) = 2. f(3+) = x - 1 = 2. f(3-) =\frac{(x+3)}{3} =\frac{6}{3} = 2.\text{ Hence continuous.}$
B. $f(3) = 4. f(3+) = f(3-) = 8 - 3 = 5. \text{ So, not continuous.}$
C. $f(3) = f(3-) = x + 3 = 6. f(3+) = x - 4 = -1.\text{ So, not continuous.}$
D. $f(3)\text{ is not existing. So, not continuous.}$
Correct Answer: $A$