Question

Evaluate the limit: \[ \lim_{x \to 2} \frac{(x^3 - 8) \sin(x - 2)}{x^2 - 4x + 4} \]

• A. 2
• B. 3
• C. 4
• D. 5

Hint:

For limits involving \( \sin(x - a) \), apply the standard limit property \( \lim_{y \to 0} \frac{\sin y}{y} = 1 \).

Answer 1

The Correct Option is B

First, notice that the denominator simplifies to \((x - 2)^2\): \[ x^2 - 4x + 4 = (x - 2)^2 \] So the limit becomes: \[ \lim_{x \to 2} \frac{(x - 2)(x^2 + 2x + 4) \sin(x - 2)}{(x - 2)^2} \] Canceling one \((x - 2)\) from both the numerator and denominator, we get: \[ \lim_{x \to 2} \frac{(x^2 + 2x + 4) \sin(x - 2)}{x - 2} \] Using the standard limit \(\lim_{y \to 0} \frac{\sin y}{y} = 1\), we evaluate the limit as: \[ (x^2 + 2x + 4) \quad \text{at} \quad x = 2 \quad \Rightarrow \quad 2^2 + 2(2) + 4 = 12 \] Thus, the limit is: \[ \lim_{x \to 2} \frac{(x^2 + 2x + 4) \sin(x - 2)}{x - 2} = 12 \]