Nilai \( \displaystyle \lim_{x \to 1} \ \frac{(x^2+x-2) \sin(x^2-1)}{x^2-2x+1} = \cdots \)
- \( -4 \)
- \( -\frac{3}{4} \)
- \( \frac{1}{2} \)
- 3
- 6
(UM UNDIP 2011)
Pembahasan:
\begin{aligned} \lim_{x \to 1} \ \frac{(x^2+x-2) \sin(x^2-1)}{x^2-2x+1} &= \lim_{x \to 1} \ \frac{(x+2)(x-1) \sin(x+1)(x-1)}{(x-1)(x-1)} \\[8pt] &= \lim_{x \to 1} \ \frac{(x+2)\sin(x+1)(x-1)}{(x-1)} \\[8pt] &= \lim_{x \to 1} \ (x+2) \cdot \lim_{x \to 1} \ \frac{\sin((x+1)(x-1))}{(x-1)} \\[8pt] &= \lim_{x \to 1} (x+2) \cdot (x+1) \\[8pt] &= (1+2)(1+1) = 3 \cdot 2 = 6 \end{aligned}
Jawaban E.