Evaluate \(\lim \limits_{x \to 3} \frac{x^{2} - 2x - 3}{x - 3}\)
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Correct Answer: Option A
Explanation:
\(\lim \limits_{x \to 3} \frac{x^{2} - 2x - 3}{x - 3}\)
\(\frac{x^{2} - 2x - 3}{x - 3} = \frac{x^{2} - 3x + x - 3}{x - 3}\)
\(\frac{(x - 3)(x + 1)}{x - 3} = x + 1\)
\(\lim \limits_{x \to 3} \frac{x^{2} - 2x - 3}{x - 3} \equiv \lim \limits_{x \to 3} (x + 1)\) (L'Hopital rule)
\(\lim \limits_{x \to 3} (x + 1) = 3 + 1 = 4\)
\(\lim \limits_{x \to 3} \frac{x^{2} - 2x - 3}{x - 3}\)
\(\frac{x^{2} - 2x - 3}{x - 3} = \frac{x^{2} - 3x + x - 3}{x - 3}\)
\(\frac{(x - 3)(x + 1)}{x - 3} = x + 1\)
\(\lim \limits_{x \to 3} \frac{x^{2} - 2x - 3}{x - 3} \equiv \lim \limits_{x \to 3} (x + 1)\) (L'Hopital rule)
\(\lim \limits_{x \to 3} (x + 1) = 3 + 1 = 4\)