Question 22

A function $f(x)$ is continuous at $x = 3$ if and only if $\lim_{x \to 3^-} f(x) = \lim_{x \to 3^+} f(x) = f(3)$.

Let us test the options:

• Option (A):

  • $f(3) = 2$
  • Right-hand limit: $\lim_{x \to 3^+} f(x) = \lim_{x \to 3^+} (x - 1) = 3 - 1 = 2$
  • Left-hand limit: $\lim_{x \to 3^-} f(x) = \lim_{x \to 3^-} \frac{x + 3}{3} = \frac{3 + 3}{3} = \frac{6}{3} = 2$
    Since $\text{LHL} = \text{RHL} = f(3) = 2$, function A is continuous at $x = 3$.

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

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

• Option (D): As $x \to 3$, $x^3 - 27 \to 0$, so $\lim_{x \to 3} \frac{1}{x^3 - 27}$ is infinite (infinite discontinuity).

Thus, only function A is continuous at $x = 3$.