Nilai \( \displaystyle \lim_{x \to 1} \ \frac{x^2-5x+4}{x^3-1} = \cdots \)
- 3
- \( 2 \frac{1}{2} \)
- 2
- 1
- -1
(UN SMA IPA 2007)
Pembahasan:
\begin{aligned} \lim_{x \to 1} \ \frac{x^2-5x+4}{x^3-1} &= \lim_{x \to 1} \ \frac{(x-1)(x-4)}{(x-1)(x^2+x+1)} \\[8pt] &= \lim_{x \to 1} \ \frac{(x-4)}{(x^2+x+1)} \\[8pt] &= \frac{1-4}{1^2+1+1} = \frac{-3}{3} \\[8pt] &= -1 \end{aligned}
Jawaban E.