Question

n趋于正无穷，lim[(n^2+1)^(1/2)/(n+1)]^n

Answer 1

lim[(n^2+1)^(1/2)/(n+1)]^n=lim{[(n²+1)/(n+1)²])^(1/2)]}^n=lim[(n²+1)/(n²+2n+1)]^n/2
=lim[(n²+2n+1-2n)/(n²+2n+1)]^n/2=lim[1-2n/(n²+2n+1)]^[-(n²+2n+1)/2n×(-2n)/(n²+2n+1)]×n/2]
n趋于正无穷,，则2n/(n²+2n+1）趋于0，(n²+2n+1)/2n趋于正无穷
lim(-2n)/(n²+2n+1)]×n/2=lim[-n²/(n²+2n+1)]=-1
所以n趋于正无穷时lim[1-2n/(n²+2n+1)]^[-(n²+2n+1)/2n×(-2n)/(n²+2n+1)]×n/2]=e^-1=1/e

Answer 2

lim[(n^2+1)^(1/2)/(n+1)]^n
=lim[1+(n^2+1)^(1/2)/(n+1)-1]^n
=lim{[1+(n^2+1)^(1/2)/(n+1)-1]^(1/[(n^2+1)^(1/2)/(n+1)-1]}^n[n^2+1)^(1/2)/(n+1)-1]
指数[1+(n^2+1)^(1/2)/(n+1)-1]^(1/[(n^2+1)^(1/2)/(n+1)-1]趋于e
指数n[n^2+1)^(1/2)/(n+1)-1]
=(n/(n+1)*(n^2+1)^(1/2)-n-1)
=(n/(n+1)*(n^2+1-n^2-1-2n)/((n^2+1)^(1/2)+n+1)
==(n/(n+1)* (-2n)/[(n^2+1)^(1/2)+n+1] (分子分母除以n)
趋于-1
极限为1/e