Which one of the following functions is continuous at $x = 3$?

A function $f(x)$ is continuous at $x = 3$ if and only if:

Let's test the options:

• Option (A):

  • Left-hand limit: $\lim_{x \to 3^-} f(x) = \lim_{x \to 3^-} \frac{x+3}{3} = \frac{3+3}{3} = 2$

  • Right-hand limit: $\lim_{x \to 3^+} f(x) = \lim_{x \to 3^+} (x-1) = 3 - 1 = 2$

  • Function value: $f(3) = 2$

  • Since $\lim_{x \to 3^-} f(x) = \lim_{x \to 3^+} f(x) = f(3) = 2$, $f(x)$ is continuous at $x = 3$.

• Option (B): $\lim_{x \to 3} f(x) = \lim_{x \to 3} (8-x) = 5 \neq f(3) = 4$ (Discontinuous).

• Option (C): $\lim_{x \to 3^-} f(x) = 3 + 3 = 6$, while $\lim_{x \to 3^+} f(x) = 3 - 4 = -1$ (Discontinuous).

• Option (D): $f(3)$ is undefined, and $\lim_{x \to 3} f(x)$ does not exist (Discontinuous).

Hence, the correct option is (A).