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Use L'Hôpital's rule: lim_(x→0) (sin x - x)/x³ =
A-1/6
B0
C1/6
D-1
Answer & Solution
Correct answer: A. -1/6
0/0 form. Differentiate top/bottom: (cos x - 1)/(3x²) — still 0/0. Again: (-sin x)/(6x) — still 0/0. Again: (-cos x)/6 → -1/6. (Or use Taylor: sin x = x - x³/6 + ..., so (sin x - x)/x³ → -1/6.)
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