Question

求助！！！极限问题求详解！

求助！！！极限问题求详解！大学数学 向左转|向右转

Answer 1

向左转|向右转
解：
x(n+1)=3+4/xn=(3xn+4)/xn
x(n+1)+1=(3xn+4)/xn
+1=(3xn+4+xn)/xn=4(xn+1)/xn
x(n+1)-4=(3xn+4)/xn
-4=(3xn+4
-4xn)/xn=-(xn-4)/xn
x1=4时，x(n+1)-4=0，x(n+1)=4
数列{xn}是各项均为4的常数数列。
lim
xn
 
=4
n→+∞
x1>0且x1≠4时，
[x(n+1)+1]/[x(n+1)-4]=-4(xn+1)/(xn-4)
[x(n+1)+1]/[x(n+1)-4]
/
(xn+1)/(xn-4)=-4，为定值
数列{(xn+1)/(xn-4)}是以(x1+1)/(x1-4)为首项，-4为公比的等比数列
(xn+1)/(xn-4)=[(x1+1)/(x1-4)]·(-4)ⁿ⁻¹
{[(x1+1)/(x1-4)]·(-4)ⁿ⁻¹-1}xn=1-[(x1+1)/(x1-4)]·(-4)ⁿ
xn={1-[(x1+1)/(x1-4)]·(-4)ⁿ}/{[(x1+1)/(x1-4)]·(-4)ⁿ⁻¹-1}
=[(x1-4)-(x1+1)·(-4)ⁿ]/[(x1+1)·(-4)ⁿ⁻¹-(x1-4)]
=4[(x1-4)-(x1+1)·(-4)ⁿ]/[(x1+1)·(-4)ⁿ-4(x1-4)]
limxn
=4[(x1-4)-(x1+1)·(-4)ⁿ]/[(x1+1)·(-4)ⁿ-4(x1-4)]
n→+∞
limxn
=4[(x1-4)/(-4)ⁿ
-(x1+1)]/[(x1+1)
-4(x1-4)/(-4)ⁿ]
n→+∞
=4[0-(x1+1)]/(x1+1-0)
=-4
综上，得：
x1=4时，
lim
xn
 
=4
n→+∞
x1>0且x1≠4时，
lim
xn
 
=-4
n→+∞