Nilai \( \displaystyle \lim_{x \to 2} \ \left( \frac{2}{x^2-4} - \frac{3}{x^2+2x-8} \right ) = \cdots \)
- \( -\frac{7}{12} \)
- \( -\frac{1}{4} \)
- \( -\frac{1}{12} \)
- \( -\frac{1}{24} \)
- \( 0 \)
(UN SMA IPA 2004)
Pembahasan:
\begin{aligned} \lim_{x \to 2} \ \left( \frac{2}{x^2-4} - \frac{3}{x^2+2x-8} \right) &= \lim_{x \to 2} \ \frac{2(x^2+2x-8)-3(x^2-4)}{(x^2-4)(x^2+2x-8)} \\[8pt] &= \lim_{x \to 2} \ \frac{2x^2+4x-16-3x^2+12}{(x^2-4)(x^2+2x-8)} \\[8pt] &= \lim_{x \to 2} \ \frac{-x^2 + 4x-4}{(x^2-4)(x^2+2x-8)} \\[8pt] &= \lim_{x \to 2} \ \frac{-(x-2)(x-2)}{(x-2)(x+2)(x+4)(x-2)} \\[8pt] &= \lim_{x \to 2} \ \frac{-1}{(x+2)(x+4)} \\[8pt] &= \frac{-1}{(2+2)(2+4)} = -\frac{1}{24} \end{aligned}
Jawaban D.