Question

用定义证明数列√(n+1)-√n的极限是0

Answer 1

|√(n+1)-√n -0| < ε

|1/(√(n+1) + √n )| < ε

1/(2√n) < ε

n > { 1/(2ε) }^2

∀ε>0 ,∃N = [{ 1/(2ε) }^2] +1, st

|√(n+1)-√n -0| < ε , ∀N>n

=>

lim(n->∞) [√(n+1)-√n]=0