Let \(f(x) = \displaystyle\frac{x^{2} - 9}{x^{2} - x - 6}\), \(x \ne 3\), \(x \ne -2\). Consider the following statements:
(A) \(\displaystyle\lim_{x \to 3} f(x) = \frac{6}{5}\)
(B) f is continuous at \(x = 0\).
(C) \(\displaystyle\lim_{x \to -2} f(x)\) exists as a finite number.
(D) If f(3) is defined as 1, then f becomes continuous at \(x = 3\).
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A) and (B) only
Factorising the numerator and the denominator:
\( \displaystyle f(x) = \frac{(x - 3)(x + 3)}{(x - 3)(x + 2)} \)
\( \displaystyle = \frac{x + 3}{x + 2}, \quad x \ne 3,\ x \ne -2 \)
Near x = −2 the numerator tends to 1 while the denominator tends to 0.
So the limit at 3 is \(\frac{6}{5}\), f is a quotient of polynomials with non-zero denominator at 0, and f becomes unbounded near −2.
Checking the statements
(A) \(\lim_{x \to 3} f(x) = \frac{6}{5}\) — Correct. \(\frac{3 + 3}{3 + 2} = \frac{6}{5}\).
(B) f is continuous at \(x = 0\). — Correct. The denominator at 0 is −6 ≠ 0, so f(0) = 3/2 equals the limit.
(C) \(\lim_{x \to -2} f(x)\) exists as a finite number. — Incorrect. \(\frac{x + 3}{x + 2}\) has numerator near 1 and denominator near 0, so it is unbounded.
(D) If f(3) is defined as 1, then f becomes continuous at \(x = 3\). — Incorrect. The limit at 3 is \(\frac{6}{5}\), not 1.