Nilai \( \displaystyle \lim_{x \to 3} \ \frac{\sin (2x-6)}{\sqrt{4-x}-1} = \cdots \)
- 4
- 2
- 0
- -2
- -4
(SBMPTN 2018)
Pembahasan:
\begin{aligned} \lim_{x \to 3} \ \frac{\sin (2x-6)}{\sqrt{4-x}-1} &= \lim_{x \to 3} \ \frac{\sin (2x-6)}{\sqrt{4-x}-1} \times \frac{\sqrt{4-x}+1}{\sqrt{4-x}+1} \\[8pt] &= \lim_{x \to 3} \ \frac{(\sqrt{4-x}+1)\sin (2x-6)}{(4-x)-1} \\[8pt] &= \lim_{x \to 3} \ \frac{(\sqrt{4-x}+1)\sin 2(x-3)}{-(x-3)} \\[8pt] &= \lim_{x \to 3} \ \frac{(\sqrt{4-x}+1)}{-1} \cdot \lim_{x \to 3} \ \frac{\sin 2(x-3)}{(x-3)} \\[8pt] &= \frac{(\sqrt{4-3} + 1)}{-1} \cdot 2 = \frac{\sqrt{1}+1}{-1} \cdot 2 \\[8pt] &= -4 \end{aligned}
Jawaban E.