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