Nilai \( \displaystyle \lim_{x \to 2} \ \frac{x^2-2x}{\sqrt{x^2+5}-3} = \cdots \)
- -6
- -3
- 0
- 3
- 6
(UMB PTN 2013)
Pembahasan:
\begin{aligned} \lim_{x \to 2} \ \frac{x^2-2x}{\sqrt{x^2+5}-3} &= \lim_{x \to 2} \ \frac{x^2-2x}{\sqrt{x^2+5}-3} \times \frac{\sqrt{x^2+5}+3}{\sqrt{x^2+5}+3} \\[8pt] &= \lim_{x \to 2} \ \frac{(x^2-2x)(\sqrt{x^2+5}+3)}{(x^2+5)-9} \\[8pt] &= \lim_{x \to 2} \ \frac{x(x-2)(\sqrt{x^2+5}+3)}{(x-2)(x+2)} \\[8pt] &= \lim_{x \to 2} \ \frac{x(\sqrt{x^2+5}+3)}{(x+2)} = \frac{2(\sqrt{2^2+5}+3)}{(2+2)} \\[8pt] &= \frac{2(\sqrt{9}+3)}{4} = \frac{12}{4} = 3 \end{aligned}
Jawaban D.