求极限 过程详细点
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/09/13 23:23:02
求极限 过程详细点
利用
(e^x)-1 x,1-cosx x²/2 (x→0),
可得
g.e.= lim(x→0)[e^(2-2cosx)]*lim(x→0){[e^(x²-2+2cosx)]-1}/(x^4)
= 1*lim(x→0)(x²-2+2cosx)/(x^4) (0/0)
= lim(x→0)(2x-0-2sinx)/(4x³) (0/0)
= lim(x→0)(2-2cosx)/(12x²)
= (1/6)*lim(x→0)(1-cosx)/x²
= (1/6)*(1/2)
= 1/12
(e^x)-1 x,1-cosx x²/2 (x→0),
可得
g.e.= lim(x→0)[e^(2-2cosx)]*lim(x→0){[e^(x²-2+2cosx)]-1}/(x^4)
= 1*lim(x→0)(x²-2+2cosx)/(x^4) (0/0)
= lim(x→0)(2x-0-2sinx)/(4x³) (0/0)
= lim(x→0)(2-2cosx)/(12x²)
= (1/6)*lim(x→0)(1-cosx)/x²
= (1/6)*(1/2)
= 1/12