Nilai \( \displaystyle \lim_{x \to -4} \ \frac{1-\cos(x+4)}{x^2+8x+16} = \cdots \)
- \( -2 \)
- \( -\frac{1}{2} \)
- \( \frac{1}{3} \)
- \( \frac{1}{2} \)
- \( 2 \)
(UM UGM 2017)
Pembahasan:
\begin{aligned} \lim_{x \to -4} \ \frac{1-\cos(x+4)}{x^2+8x+16} &= \lim_{x \to -4} \ \frac{2\sin^2 \frac{1}{2}(x+4)}{(x+4)(x+4)} \\[8pt] &= 2 \cdot \lim_{x \to -4} \ \frac{\sin \frac{1}{2}(x+4)}{(x+4)} \cdot \lim_{x \to -4} \ \frac{\sin \frac{1}{2}(x+4)}{(x+4)} \\[8pt] &= 2 \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{2} \end{aligned}
Jawaban D.