Question

an=根号n+2(根号（n+1）-根号（n-1）),求数列an的极限

Answer 1

an=√(n+2)*[√(n+1)-√(n-1)]

=√(n+2)*2/[√(n+1)+√(n-1)] (分子有理化)

=2/{√[(n+1)/(n+2)]+√[(n-1)/(n+2)]} (分子分母同除以√(n+2))

所以,n→∞时,an→2/(1+1)=1,

即 lim(n→∞)an=1.