Question

求极限，答案e/2，我算出来是-e/2

Answer 1

楼主应该是泰勒展开的时候少展开了一项,才会得到e的,确实应该是e/2.
(1+1/n)^(n+1)=e^((n+1)*ln(1+1/n)),
泰勒展开,ln(1+1/n)=1/n-1/2*1/n^2+o(1/n^2).
故(n+1)*ln(1+1/n)
=(n+1)(1/n-1/2*1/n^2+o(1/n^2))
=1-1/2*1/n+1/n+o(1/n)
=1+1/2*1/n+o(1/n)
所以
(1+1/n)^(n+1)-e=e^(1+1/2*1/n+o(1/n))-e=e*e^(1/2*1/n+o(1/n)-1)=e*1/2*1/n+o(1/n).
所以
sin((1+1/n)^(n+1)-e)=(1+1/n)^(n+1)-e=e*1/2*1/n+o(1/n).
所以
n*sin[(1+1/n)^(n+1)-e]=e*1/2+o(1),
因此极限是确实是e/2.

追问

谢谢，我刚是用等价代换和洛必达法则做的😂😂