Nilai \( \displaystyle \lim_{x\to 0} \ \frac{\tan(2x^3)}{x^3+\sin^3 4x} = \cdots \)
- 5/65
- 4/65
- 3/65
- 2/65
- 1/65
Pembahasan:
\begin{aligned} \lim_{x\to 0} \ \frac{\tan(2x^3)}{x^3+\sin^3 4x} &= \lim_{x\to 0} \ \frac{\tan(2x^3)}{x^3+\sin^3 4x} \times \frac{\frac{1}{x^3}}{\frac{1}{x^3}} \\[8pt] &= \lim_{x\to 0} \ \frac{ \frac{\tan(2x^3)}{x^3}}{1+\frac{\sin^3 4x}{x^3}} \\[8pt] &= \frac{\displaystyle \lim_{x\to 0} \ \frac{\tan(2x^3)}{x^3}}{\displaystyle \lim_{x\to 0} \ 1 + \lim_{x\to 0} \ \frac{\sin^3 4x}{x^3}} \\[8pt] &= \frac{2}{1+4^3} = \frac{2}{1+64}= \frac{2}{65} \end{aligned}
Jawaban D.