Nilai \( \displaystyle \lim_{x \to 2} \ \frac{x^2-6x+8}{3-\sqrt{17-2x^2}} = \cdots \)
- 6
- 3/2
- 0
- -3/2
- -6
(UNBK SMA IPA 2019)
Pembahasan:
\begin{aligned} \lim_{x \to 2} \ \frac{x^2-6x+8}{3-\sqrt{17-2x^2}} &= \lim_{x \to 2} \ \frac{x^2-6x+8}{3-\sqrt{17-2x^2}} \times \frac{3+\sqrt{17-2x^2}}{3+\sqrt{17-2x^2}} \\[8pt] &= \lim_{x \to 2} \ \frac{(x-2)(x-4)(3+\sqrt{17-2x^2})}{9-(17-2x^2)} \\[8pt] &= \lim_{x \to 2} \ \frac{(x-2)(x-4)(3+\sqrt{17-2x^2})}{2x^2-8} \\[8pt] &= \lim_{x \to 2} \ \frac{(x-2)(x-4)(3+\sqrt{17-2x^2})}{2(x-2)(x+2)} \\[8pt] &= \lim_{x \to 2} \ \frac{(x-4)(3+\sqrt{17-2x^2})}{2(x+2)} \\[8pt] &= \frac{(2-4)(3+\sqrt{17-2(2)^2})}{2(2+2)} \\[8pt] &= \frac{(-2)(3+\sqrt{9})}{8} = -\frac{3}{2} \end{aligned}
Jawaban D.