Question

求证lim（1+1/n+1/n2）n =e ( n→∞)

式中的2是平方！！

Answer 1

ln{[1+1/n+1/(n^2)]^n}
=n*ln[1+1/n+1/(n^2)]
=ln[1+1/n+1/(n^2)] / (1/n)

当n->∞时
ln[1+1/n+1/(n^2)]->0
1/n->0
上式的分子分母同时趋近于零。可以利用罗比他法则

lim_{n->∞}{ln[1+1/n+1/(n^2)] / (1/n)}
=lim_{n->∞}{ln[1+1/n+1/(n^2)]' / (1/n)'}('表示求导数)
=lim_{n->∞}{ [(-1/(n^2)-2/(n^3))/(1+1/n+1/(n^2))] / [-1/(n^2)] }
=lim_{n->∞}{ (n^3+2)/(n^3+n^2+n) }
=1

即
lim_{n->∞}{ln{[1+1/n+1/(n^2)]^n}}=1
lim_{n->∞}{[1+1/n+1/(n^2)]^n}=e

Answer 2

设y=(1+1/n+1/n^2)^n=[1+(n+1)/n^2]^n
所以lny=nln[1+(n+1)/n^2]
=[n*(n+1)/n^2]ln[1+(n+1)/n^2]^[n^2/(n+1)]
=[(n^2+n)/n^2]ln[1+(n+1)/n^2]^[n^2/(n+1)]
利用lim(1+1/x)^x=e,x趋向无穷大，
limlny=1*lne=1，所以limy=e( n→∞)

Answer 3

1,lim(1/n+1)与lim(1/n2)都是0，所以lim1是1

Answer 4

lim（1+1/n+1/n2）n=lim e(n ln(1+1/n+1/n2) )
lim(n+1/n)n =e
lim e (n ln(n+1/n) )=e
所以求证

Answer 5

？？题目表达不清楚，