Nilai \( \displaystyle \lim_{x \to 1} \ \frac{(x^2-1)\tan (2x-2)}{\sin^2 (x-1)} = \cdots \)
- 1
- 2
- 4
- 6
- 8
Pembahasan:
\begin{aligned} \lim_{x \to 1} \ \frac{(x^2-1)\tan (2x-2)}{\sin^2 (x-1)} &= \lim_{x \to 1} \ \frac{(x-1)(x+1) \tan 2(x-1)}{\sin^2 (x-1)} \\[8pt] &= \lim_{x \to 1} \ (x+1) \cdot \lim_{x \to 1} \ \frac{(x-1)}{\sin (x-1)} \cdot \lim_{x \to 1} \ \frac{\tan 2(x-1)}{\sin (x-1)} \\[8pt] &= (1+1) \cdot 1 \cdot 2 \\[8pt] &= 4 \end{aligned}
Jawaban C.