Nilai \( \displaystyle \lim_{x \to 2} \ \frac{\tan (2-\sqrt{2x})}{x^2 - 2x} = \cdots \)
- \( \frac{1}{4} \)
- \( \frac{1}{8} \)
- \( 0 \)
- \( -\frac{1}{6} \)
- \( -\frac{1}{4} \)
(SPMB 2005)
Pembahasan:
\begin{aligned} \lim_{x \to 2} \ \frac{\tan (2-\sqrt{2x})}{x^2 - 2x} &= \lim_{x \to 2} \ \frac{\tan (-\sqrt{2} \ (\sqrt{x}-\sqrt{2}))}{ x \ (\sqrt{x}+\sqrt{2})(\sqrt{x}-\sqrt{2}) } \\[8pt] &= \lim_{x \to 2} \ \frac{\tan (-\sqrt{2} \ (\sqrt{x}-\sqrt{2}))}{(\sqrt{x}-\sqrt{2}) } \cdot \lim_{x \to 2} \ \frac{1}{ x \ (\sqrt{x}+\sqrt{2})} \\[8pt] &= -\sqrt{2} \cdot \frac{1}{2(\sqrt{2} + \sqrt{2})} = -\sqrt{2} \cdot \frac{1}{4\sqrt{2}} \\[8pt] &= - \frac{1}{4} \end{aligned}
Jawaban E.