Discussion
\(\displaystyle\lim_{x \to 2} \frac{(x^{3} - 8)\sin(x - 2)}{x^{2} - 4x + 4}\) is equal to
Answer: C
The limit is 12.
Factor the numerator and denominator: \(x^{3} - 8 = (x - 2)(x^{2} + 2x + 4)\) and \(x^{2} - 4x + 4 = (x - 2)^{2}\). Then
\[ \begin{aligned} \frac{(x^{3} - 8)\sin(x - 2)}{(x - 2)^{2}} &= (x^{2} + 2x + 4)\frac{\sin(x - 2)}{x - 2} \\ &\to 12 \times 1 = 12 \end{aligned} \]
8 and 4 use only part of \(x^{2} + 2x + 4\) at \(x = 2\), and the negative values come from a sign slip.
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