Question

Evaluate $ \lim_{x \to 3} \frac{x^2 - x - 6}{x - 3} $.

• A. 4
• B. 5
• C. 6
• D. 7

Hint:

For limits yielding \( \frac{0}{0} \), factorize or simplify the expression to cancel common terms, then evaluate the limit directly.

Answer 1

The Correct Option is B

Evaluate the limit: \[ \lim_{x \to 3} \frac{x^2 - x - 6}{x - 3} \] Substituting \( x = 3 \) gives \( \frac{0}{0} \), so factorize the numerator: \[ x^2 - x - 6 = (x - 3)(x + 2) \] \[ \frac{x^2 - x - 6}{x - 3} = \frac{(x - 3)(x + 2)}{x - 3} = x + 2 \quad (x \neq 3) \] Now take the limit: \[ \lim_{x \to 3} (x + 2) = 3 + 2 = 5 \] Thus, the value of the limit is: \[ \boxed{5} \]