Nilai \( \displaystyle \lim_{x \to 1} \ \frac{\tan (x-1) \sin(1-\sqrt{x})}{x^2-2x+1} = \cdots \)
- -1
- \( -\frac{1}{3} \)
- 0
- \( -\frac{1}{2} \)
- 1
(UM UGM 2004)
Pembahasan:
\begin{aligned} \lim_{x \to 1} \ \frac{\tan (x-1) \sin(1-\sqrt{x})}{x^2-2x+1} &= \lim_{x \to 1} \ \frac{\tan (x-1) \sin(1-\sqrt{x})}{-(x-1)(1-x)} \\[8pt] &= \lim_{x \to 1} \ \frac{\tan (x-1) \sin(1-\sqrt{x})}{-(x-1)(1+\sqrt{x})(1-\sqrt{x})} \\[8pt] &= \lim_{x \to 1} \ \frac{1}{-(1+\sqrt{x})} \cdot \lim_{x \to 1} \ \frac{\tan(x-1)}{(x-1)} \cdot \lim_{x \to 1} \ \frac{\sin(1-\sqrt{x})}{(1-\sqrt{x})} \\[8pt] &= \frac{1}{-(1+\sqrt{1})} \cdot 1 \cdot 1 = -\frac{1}{2} \end{aligned}
Jawaban D.