Nilai \( \displaystyle \lim_{x \to -2} \ \frac{1-\cos(x+2)}{x^2+4x+4} = \cdots \)
- 0
- 1/4
- 1/2
- 2
- 4
(UMPTN 1995)
Pembahasan:
\begin{aligned} \lim_{x \to -2} \ \frac{1-\cos(x+2)}{x^2+4x+4} &= \lim_{x \to -2} \ \frac{2\sin^2 \frac{1}{2}(x+2)}{(x+2)^2} \\[8pt] &= 2 \cdot \left(\frac{1}{2} \right)^2 = 2 \cdot \frac{1}{4} \\[8pt] &= \frac{1}{2} \end{aligned}
Jawaban C.