limx→0sinx−(a+bx+cx2)x3=d(d≠0)\lim_{x \to 0} \frac{\sin x - (a + bx + cx^2)}{x^3} = d \qquad (d \neq 0)limx→0x3sinx−(a+bx+cx2)=d(d=0) =limx→0(x−x33!+x55!−⋯ )−(a+bx+cx2)x3= \lim_{x \to 0} \frac{\left(x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} - \cdots\right) - (a + bx + cx^2)}{x^3}=limx→0x3(x−3!x3+5!x5−⋯)−(a+bx+cx2) これから a=0a = 0a=0 x−bx=0x - bx = 0x−bx=0 cx2=0cx^2 = 0cx2=0 よって a=0a = 0a=0、b=1b = 1b=1、c=0c = 0c=0 このとき d=−13!=−16d = -\frac{1}{3!} = -\frac{1}{6}d=−3!1=−61