Nilai lim_(x→0)⁡ (2x sin⁡ 3x)/(1-cos⁡ 6x)=⋯

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Bahas Soal Matematika   »   Limit   ›  

Nilai \( \displaystyle \lim_{x \to 0} \ \frac{2x \sin 3x}{1-\cos 6x} = \cdots \)

  1. -1
  2. -1/3
  3. 0
  4. 1/3
  5. 1

(UN SMA IPA 2007)

Pembahasan:

\begin{aligned} \lim_{x \to 0} \ \frac{2x \sin 3x}{1-\cos 6x} &= \lim_{x \to 0} \ \frac{2x \sin 3x}{1-\cos 2(x)} \\[8pt] &= \lim_{x \to 0} \ \frac{2x \sin 3x}{1-(\cos^2 3x - \sin^2 3x)} \\[8pt] &= \lim_{x \to 0} \ \frac{2x \sin 3x}{1-(1 - \sin^2 3x - \sin^2 3x)} \\[8pt] &= \lim_{x \to 0} \ \frac{2x \sin 3x}{2 \sin^2 3x} = \lim_{x \to 0} \ \frac{x}{\sin 3x} \\[8pt] &= \frac{1}{3} \end{aligned}

Jawaban D.