求极限:limx→2/1ⅹ2sim2x limx无穷2的x次方的极限
更新时间:2021-11-08 19:04:21 • 作者:LESLIE •阅读 5967
limx趋向于2时(x 1)/(x-2)的极限
limx趋向于2时(x-1)/(x-2)的极限:
根据无穷小与无穷大的关系求解
limx趋向于2时f(x)=(x-2)/(x-1)=0/1=0,函数f(x)为当x趋于2时的无穷小
于是limx趋向于2时1/f(x)=(x-1)/(x-2)的极限:∞
X趋向于无穷大时(1+x/2)2/x的极限怎么求??
(1+x/2)2/x的极限 : 1
(1+x/2)2/x=2/x + 1 ,当X趋向于无穷大时,极限=1
Ln x dx的不定积分: ∫lnx dx = lnx
limx→2(x/2)^1/(x-2)
L=lim(x->2) (x/2)^1/(x-2)
lnL = lim(x->2) ln(x/2) /(x-2) (0/0)
=lim(x->2) (1/x)
=1/2
L = e^(1/2)
ie
lim(x->2) (x/2)^1/(x-2) =e^(1/2)
求极限limx→∞x2(1-cos1x)
利用等价无穷小:x→0时,1-cosx~
1
2 x2.
故原式
lim
x→∞
1?cos
1
x
1
x2 =
lim
x→∞
1
2
1
x2
1
x2 =
1
2 .