Question

高等数学 求极限

Answer 1

这是无穷大 - 无穷大型，可以进塌埋颤行转换
[n(n+2)]^1/2] - (n^2+1)^1/2
= {[n(n+2)]^1/2] - (n^2+1)^1/2} {[n(n+2)]^1/2] -+(n^2+1)^1/2} / { [n(n+2)]^1/2] + (n^2+1)^1/2}
= [n(n+2) - (n^2+1)] / { [n(n+2)]^1/2] + (n^2+1)^1/2}
=(n^2+2n-n^2-1) / { [n(n+2)]^1/2] + (n^2+1)^1/2}
=(2n-1)/ / { [n(n+2)]^1/2] + (n^2+1)^1/2} (分子分母同时除以n)
= (2-1/n) { [1+2/n)]^1/2] + (1+1/n^2)^1/2}
当n趋于无穷大液培时，1/n, 2/n, 1/n^2趋于0, 因此团败原极限＝2/(1+1)=1

Answer 2

lim(n->∞) { √[n(n+2)] -√(n^2+1) }
=lim(n->∞) { [n(n+2)] -(n^2+1) } / { √[n(n+2)] +√(n^2+1) }
=lim(n->∞) (2n-1) / { √升卖[n(n+2)] +√(n^2+1) }
分子分母悉笑雀同时除以 n
=lim(n->睁早∞) (2- 1/n) / [ √(1+2/n) +√(1+1/n^2) ]
=(2-0)/(1+1)
=1