limx→0ex−sinx−cosx+Ax2+Bx3x6\lim_{x \to 0} \frac{e^x - \sin x - \cos x + Ax^2 + Bx^3}{x^6}limx→0x6ex−sinx−cosx+Ax2+Bx3 =limx→0(1+x+x22!+x33!)−(x−x33!+x55!−⋯ )−(1−x22!+x44!−⋯ )+Ax2+Bx3x6= \lim_{x \to 0} \frac{\left(1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}\right) - \left(x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} - \cdots\right) - \left(1 - \dfrac{x^2}{2!} + \dfrac{x^4}{4!} - \cdots\right) + Ax^2 + Bx^3}{x^6}=limx→0x6(1+x+2!x2+3!x3)−(x−3!x3+5!x5−⋯)−(1−2!x2+4!x4−⋯)+Ax2+Bx3 =limx→0{2(x22!+x33!)+2(x66!+x77!)+⋯ }+Ax2+Bx3x6= \lim_{x \to 0} \frac{\left\{2\left(\dfrac{x^2}{2!}+\dfrac{x^3}{3!}\right) + 2\left(\dfrac{x^6}{6!}+\dfrac{x^7}{7!}\right) + \cdots\right\} + Ax^2 + Bx^3}{x^6}=limx→0x6{2(2!x2+3!x3)+2(6!x6+7!x7)+⋯}+Ax2+Bx3 有限確定であるためには 2(x22!+x33!)+Ax2+Bx3=02\left(\frac{x^2}{2!}+\frac{x^3}{3!}\right) + Ax^2 + Bx^3 = 02(2!x2+3!x3)+Ax2+Bx3=0 となればよいので ∴A=−1B=−13\therefore A = -1 \qquad B = -\frac{1}{3}∴A=−1B=−31 このとき極限値は 2⋅16!=13602 \cdot \frac{1}{6!} = \frac{1}{360}2⋅6!1=3601