Nilai \( \displaystyle \lim_{x \to 0} \ \frac{x \tan x}{x \sin x - \cos x + 1} = \cdots \)
- 2
- 3/2
- 1
- 2/3
- -1
(SBMPTN 2013)
Pembahasan:
\begin{aligned} \lim_{x \to 0} \ \frac{x \tan x}{x \sin x - \cos x + 1} &= \lim_{x \to 0} \ \frac{x \tan x}{x \sin x + 1 - \cos x} \\[8pt] &= \lim_{x \to 0} \ \frac{x \tan x}{x \sin x + 2\sin^2 \frac{1}{2}x} \cdot \frac{\frac{1}{x^2}}{\frac{1}{x^2}} \\[8pt] &= \lim_{x \to 0} \ \frac{\frac{\tan x}{x}}{\frac{\sin x}{x} + \frac{2\sin^2 \frac{1}{2}x}{x^2}} \\[8pt] &= \frac{ \displaystyle \lim_{x \to 0} \ \frac{\tan x}{x} }{ \displaystyle \lim_{x \to 0} \ \frac{\sin x}{x} + \lim_{x \to 0} \ \frac{2\sin^2 \frac{1}{2}x}{x^2} } \\[8pt] &= \frac{1}{1+ 2 \cdot (\frac{1}{2})^2} = \frac{1}{1+\frac{2}{4}} = \frac{2}{3} \end{aligned}
Jawaban D.