Question

极限运算

lim趋向于0，（e的x的平方次-e的2-2cosx次）÷x的4次方等于？

Answer 1

lim(x->0)[e^(x^2)-e^(2-2cosx)]/ x^4 (0/0)
=lim(x->0)[2xe^(x^2)-2sinx.e^(2-2cosx)]/ (4x^3)
=lim(x->0)[e^(x^2)-e^(2-2cosx)]/ (2x^2) ( sinx ~x , x->0 ) (0/0)
=lim(x->0)[2xe^(x^2)-2sinxe^(2-2cosx)]/ (4x)
=lim(x->0)[e^(x^2)-e^(2-2cosx)]/ 2
=0

追问

我也觉得是0！可是答案是12分之1。

追答

lim(x->0)[e^(x^2)/ x^4] (0/0)
=lim(x->0)[2xe^(x^2)/ (4x^3)]
=lim(x->0)[e^(x^2)/ (2x^2)] (0/0)
=lim(x->0)[2xe^(x^2)/ (4x)]
= 1/2

lim(x->0)e^(2-2cosx)/ x^4 (0/0)
= lim(x->0) sinxe^(2-2cosx)/ (2x^3) (0/0)
= lim(x->0) [cosx +2(sinx)^2] e^(2-2cosx)/(6x^2) (0/0)
=lim(x->0) [-sinx +4(sinx)cosx +2sinx(cosx+2(sinx)^2) ] e^(2-2cosx)/(12x)
=lim(x->0) [-sinx +6(sinx)cosx +4(sinx)^3 ] e^(2-2cosx)/(12x)
=lim(x->0) [-sinx +3sin(2x) +4(sinx)^3 ] e^(2-2cosx)/(12x) (0/0)
=lim(x->0){ [-cosx +6cos(2x) +12(sinx)^2cosx
+ 2sinx([-sinx +3sin(2x) +4(sinx)^3 ]) ].e^(2-2cosx) }/(12)
=(-1+6)/12
=5/12

lim(x->0)[e^(x^2)-e^(2-2cosx)]/ x^4
=1/2 -5/12
=1/12

Answer 2

写成分式，然后上下分别求导数试试。

追问

能帮我解下吗？