Nilai \( \displaystyle \lim_{x \to 2} \ \frac{\sqrt{x^2+5}-3}{x^2-2x} = \cdots \)
- 0
- 1/3
- 1/2
- 3/4
- ∞
(UM UGM 2007)
Pembahasan:
\begin{aligned} \lim_{x \to 2} \ \frac{\sqrt{x^2+5}-3}{x^2-2x} &= \lim_{x \to 2} \ \frac{\sqrt{x^2+5}-3}{x^2-2x} \times \frac{\sqrt{x^2+5}+3}{\sqrt{x^2+5}+3} \\[8pt] &= \lim_{x \to 2} \ \frac{(x^2+5)-9}{(x^2-2x)(\sqrt{x^2+5}+3)} \\[8pt] &= \lim_{x \to 2} \ \frac{(x^2-4)}{(x^2-2x)(\sqrt{x^2+5}+3)} \\[8pt] &= \lim_{x \to 2} \ \frac{(x-2)(x+2)}{x(x-2)(\sqrt{x^2+5}+3)} \\[8pt] &= \lim_{x \to 2} \ \frac{(x+2)}{x \ (\sqrt{x^2+5}+3)} = \frac{(2+2)}{2 \ (\sqrt{2^2+5}+3)} \\[8pt] &= \frac{4}{2(\sqrt{9}+3)} = \frac{4}{12} = \frac{1}{3} \end{aligned}
Jawaban B.