lim √(n^2-3n)/2n+1 n→0
人气:210 ℃ 时间:2020-04-11 17:28:21
解答
如果是数列极限,就只有n→∞,这一种极限
要趋于某数,一定要函数极限
x→0
lim √(x^2-3x) / (2x+1)
直接代入,就有
=0/(0+1)
=0
n→∞
lim √(n^2-3n) / (2n+1)
上下同时除以n
=lim √(n^2-3n)/n / (2n+1)/n
=lim √(1-(3/n)) / (2+(1/n))
因为1/n趋于0,故
=√(1-0) / (2+0)
=1/2
有不懂欢迎追问根号里的n^2-3n除以n为什么就等于根号1-3/n?√(n^2-3n)/n=√[(n^2-3n)/n^2]=√[1-(3n/n^2)]=√(1-(3/n))有不懂欢迎追问
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