Question

lim[(x^5+7x^4+2)^1/5-x]=b,求,b的值。

Answer 1

x-->∞
lim（x∧5+7x∧4+2）∧a－x＝b，b≠0.求常数阿a，b。
记p=x^5+7x^4+2
则lim[p^a-x]=b
lim[(p^a-x)/x]=0=lim[p^a/x-1]
即lim[p^a/x]=1
【p/x^(1/a)】^a=[x^(5-1/a)(1+7/x+2/x^5)]^a
此式极限为1
∴x指数
5-1/a=0
a=1/5
lim（x∧5+7x∧4+2）^(1/5)－x＝b
limx{[（1+7/x+2/x^5）]^(1/5)-1}＝b
则
lim
[（1+7/x+2/x^5）]^(1/5)-1=lim
b/x
lim
e^1/5ln
[（1+7/x）]-1=lim
b/x
lim
e^1/5*7/x-1=lim
b/x
lim
(1+7/5*1/x-1)=lim
b/x
∴
b=7/5
当x趋于0
limcotx［（1╱sinx）-（1╱tanx）］
=limcotx*1/sinx*(1-cosx)
=limcotx*2sin^(x/2)/sinx
=imcotx*sin(x/2)
=imcotx*sin(x/2)/cos(x/2)
=lim
cotx*tan(x/2)
tan(x/2)=t
==>
lim
cotx*tan(x/2)
=lim
t/[2t/(1-t^2)]
=1/2

Answer 2

简单分析一下即可，详情如图所示