Nilai \( \displaystyle \lim_{x \to 0} \ \frac{(x^2-1) \sin 6x}{x^3+3x^2+2x} = \cdots \)
- -3
- -2
- 2
- 3
- 5
(UMPTN 1995)
Pembahasan:
\begin{aligned} \lim_{x \to 0} \ \frac{(x^2-1) \sin 6x}{x^3+3x^2+2x} &= \lim_{x \to 0} \ \frac{(x+1)(x-1) \sin 6x}{x(x+1)(x+2)} \\[8pt] &= \lim_{x \to 0} \ \frac{(x-1) \sin 6x}{x(x+2)} \\[8pt] &= \lim_{x \to 0} \ \frac{(x-1)}{(x+2)} \cdot \lim_{x \to 0} \ \frac{\sin 6x}{x} \\[8pt] &= \frac{(0-1)}{(0+2)} \cdot 6 = \frac{-1}{2} \cdot 6 \\[8pt] &= -3 \end{aligned}
Jawaban A.