Nilai \( \displaystyle \lim_{x \to 1} \frac{\sqrt{2-x}-x}{x^2-x} = \cdots \)
- \( -1 \frac{1}{2} \)
- -1
- 0
- 1
- \( 1 \frac{1}{2} \)
(SPMB 2007)
Pembahasan:
\begin{aligned} \lim_{x \to 1} \ \frac{\sqrt{2-x}-x}{x^2-x} &= \lim_{x \to 1} \ \frac{\sqrt{2-x}-x}{x^2-x} \times \frac{\sqrt{2-x}+x}{\sqrt{2-x}+x} \\[8pt] &= \lim_{x \to 1} \ \frac{(2-x)-x^2}{(x^2-x)(\sqrt{2-x}+x)} \\[8pt] &= \lim_{x \to 1} \ \frac{-(x^2+x-2)}{(x^2-x)(\sqrt{2-x}+x)} \\[8pt] &= \lim_{x \to 1} \ \frac{-(x+2)(x-1)}{x(x-1)(\sqrt{2-x}+x)} \\[8pt] &= \lim_{x \to 1} \ \frac{-(x+2)}{x \ (\sqrt{2-x}+x)} \\[8pt] &= \frac{-(1+2)}{(1) \ (\sqrt{2-1}+1)} = -\frac{3}{2} = - 1 \frac{1}{2} \end{aligned}
Jawaban A.