解答
x→0+lim(x(1−cos(x))1−cos(x))
解答
21
+1
十进制
0.5求解步骤
x→0+lim(x(1−cos(x))1−cos(x))
乘开 x(1−cos(x)):x−xcos(x)
=x→0+lim(x−xcos(x)1−cos(x))
使用洛必达法则
=x→0+lim1−cos(x)+2xsin(x)2cos(x)sin(x)
化简 1−cos(x)+2xsin(x)2cos(x)sin(x):cos(x)(2−2cos(x)+xsin(x))sin(x)
=x→0+lim(cos(x)(2−2cos(x)+xsin(x))sin(x))
使用洛必达法则
=x→0+lim−2cos(x)sin(x)(2−2cos(x)+xsin(x))+2xcos(x)(3sin(x)+xcos(x))cos(x)
化简 −2cos(x)sin(x)(2−2cos(x)+xsin(x))+2xcos(x)(3sin(x)+xcos(x))cos(x):−xsin(x)(2−2cos(x)+xsin(x))+cos(x)(3sin(x)+xcos(x))2xcos(x)cos(x)
=x→0+lim(−xsin(x)(2−2cos(x)+xsin(x))+cos(x)(3sin(x)+xcos(x))2xcos(x)cos(x))
使用洛必达法则
=x→0+lim2x−3xsin(x)cos(x)−4xcos(x)+4xcos(x)cos(x)−2xxcos(x)sin(x)−10xsin(x)sin(x)−xsin(x)cos(x)−2sin(x)+2sin(x)cos(x)+4cos(x)cos(x)2(2x1cos23(x)+(−23sin(x)cos(x))x)
化简 2x−3xsin(x)cos(x)−4xcos(x)+4xcos(x)cos(x)−2xxcos(x)sin(x)−10xsin(x)sin(x)−xsin(x)cos(x)−2sin(x)+2sin(x)cos(x)+4cos(x)cos(x)2(2x1cos23(x)+(−23sin(x)cos(x))x):−3xsin(x)cos(x)−4xcos(x)+4xcos(x)cos(x)−2xxcos(x)sin(x)−10xsin(x)sin(x)−xsin(x)cos(x)−2sin(x)+2sin(x)cos(x)+4cos(x)cos(x)2(−3xsin(x)cos(x)+cos23(x))
=x→0+lim−3xsin(x)cos(x)−4xcos(x)+4xcos(x)cos(x)−2xxcos(x)sin(x)−10xsin(x)sin(x)−xsin(x)cos(x)−2sin(x)+2sin(x)cos(x)+4cos(x)cos(x)2(−3xsin(x)cos(x)+cos23(x))
代入值 x=0=−3⋅0⋅sin(0)cos(0)−4⋅0⋅cos(0)+4⋅0⋅cos(0)cos(0)−2⋅0⋅0cos(0)sin(0)−100sin(0)sin(0)−0sin(0)cos(0)−2sin(0)+2sin(0)cos(0)+4cos(0)cos(0)2(−3⋅0⋅sin(0)cos(0)+cos23(0))
化简 −3⋅0⋅sin(0)cos(0)−4⋅0⋅cos(0)+4⋅0⋅cos(0)cos(0)−2⋅0⋅0cos(0)sin(0)−100sin(0)sin(0)−0sin(0)cos(0)−2sin(0)+2sin(0)cos(0)+4cos(0)cos(0)2(−3⋅0⋅sin(0)cos(0)+cos23(0)):21
=21