Nilai \( \displaystyle \lim_{x\to 0} \ \frac{\sqrt{x}-\sqrt{\sin x}}{x^2\sqrt{x}} = \cdots \)
- 12
- 6
- 1/3
- 1/12
- 2/3
Pembahasan:
\begin{aligned} &\lim_{x\to 0} \ \frac{x-\sin x}{x^3} = \frac{1}{6} \\[8pt] &\lim_{x\to 0} \ \frac{(2x)-\sin (2x)}{(2x)^3} = \frac{1}{6} \\[8pt] &\lim_{x\to 0} \ \frac{(3x)-\sin (3x)}{(3x)^3} = \frac{1}{6} \\[8pt] \end{aligned}
\begin{aligned} \lim_{x\to 0} \ \frac{\sqrt{x}-\sqrt{\sin x}}{x^2\sqrt{x}} &= \lim_{x\to 0} \ \frac{\sqrt{x}-\sqrt{\sin x}}{x^2\sqrt{x}} \times \frac{\sqrt{x}+\sqrt{\sin x}}{\sqrt{x}+\sqrt{\sin x}} \\[8pt] &= \lim_{x\to 0} \ \frac{x-\sin x}{x^2\sqrt{x} \ (\sqrt{x}+\sqrt{\sin x})} \\[8pt] &= \lim_{x\to 0} \ \frac{x-\sin x}{x^2\sqrt{x} \cdot \sqrt{x} \ \left(1+\sqrt{ \frac{\sin x}{x}}\right)} \\[8pt] &= \lim_{x\to 0} \ \frac{x-\sin x}{x^3 \ \left(1+\sqrt{ \frac{\sin x}{x}}\right)} \\[8pt] &= \lim_{x\to 0} \ \frac{x-\sin x}{x^3} \cdot \lim_{x\to 0} \ \frac{1}{\left(1+\sqrt{ \frac{\sin x}{x}}\right)} \\[8pt] &= \frac{1}{6} \cdot \frac{1}{\left(1+\sqrt{ \displaystyle \lim_{x\to 0} \ \frac{\sin x}{x}}\right)} \\[8pt] &= \frac{1}{6} \cdot \frac{1}{1+1} = \frac{1}{12} \end{aligned}
Jawaban D.