Find \(\lim\limits_{x \to 3} \frac{2x^{2} + x - 21}{x - 3}\).
Take Free Practice Test On 2026 JAMB UTME, Post UTME, WAEC SSCE, GCE, NECO SSCE
Correct Answer: Option D
Explanation:
\(\lim\limits_{x \to 3} \frac{2x^{2} + x - 21}{x - 3}\)
\(2x^{2} + x - 21 = 2x^{2} - 6x + 7x - 21 \) (by factorizing)
= \((2x + 7)(x - 3)\)
\(\therefore \lim\limits_{x \to 3} \frac{2x^{2} + x - 21}{x - 3} \equiv \lim\limits_{x \to 3} \frac{(2x+7)(x-3)}{x-3}\)
\(\lim\limits_{x \to 3} (2x + 7) = 2(3) + 7 = 13\)
\(\lim\limits_{x \to 3} \frac{2x^{2} + x - 21}{x - 3}\)
\(2x^{2} + x - 21 = 2x^{2} - 6x + 7x - 21 \) (by factorizing)
= \((2x + 7)(x - 3)\)
\(\therefore \lim\limits_{x \to 3} \frac{2x^{2} + x - 21}{x - 3} \equiv \lim\limits_{x \to 3} \frac{(2x+7)(x-3)}{x-3}\)
\(\lim\limits_{x \to 3} (2x + 7) = 2(3) + 7 = 13\)