题目内容
15.$\underset{lim}{n→∞}$($\frac{1}{2}×\frac{3}{4}×\frac{5}{6}…\frac{2n-1}{2n}$)=0.分析 由题意知$\frac{2n-1}{2n}$<$\frac{2n}{2n+1}$,从而可得0<$\frac{1}{2}×\frac{3}{4}×\frac{5}{6}…\frac{2n-1}{2n}$<$\frac{1}{\sqrt{2n+1}}$,从而求极限.
解答 解:∵$\frac{2n-1}{2n}$<$\frac{2n}{2n+1}$,
∴($\frac{1}{2}×\frac{3}{4}×\frac{5}{6}…\frac{2n-1}{2n}$)2=$\frac{1}{2}$×$\frac{1}{2}$×$\frac{3}{4}$×$\frac{3}{4}$×…×$\frac{2n-1}{2n}$×$\frac{2n-1}{2n}$
<$\frac{1}{2}$×$\frac{2}{3}$×$\frac{3}{4}$×$\frac{4}{5}$×…×$\frac{2n-1}{2n}$×$\frac{2n}{2n+1}$=$\frac{1}{2n+1}$,
∴0<$\frac{1}{2}×\frac{3}{4}×\frac{5}{6}…\frac{2n-1}{2n}$<$\frac{1}{\sqrt{2n+1}}$,
又∵$\underset{lim}{n→∞}$0=0,$\underset{lim}{n→∞}$$\frac{1}{\sqrt{2n+1}}$=0,
∴$\underset{lim}{n→∞}$($\frac{1}{2}×\frac{3}{4}×\frac{5}{6}…\frac{2n-1}{2n}$)=0,
故答案为:0.
点评 本题考查了极限的求法.
练习册系列答案
相关题目
20.已知椭圆方程为$\frac{{x}^{2}}{8}+\frac{{y}^{2}}{2}=1$,过椭圆上一点P(2,1)作切线交y轴于N,过P的另一条直线交y轴于M,若△PMN是以MN为底边的等腰三角形,则直线PM的方程为( )
A. | y=$\frac{3}{2}x-2$ | B. | y=$\frac{1}{2}x$ | C. | y=-2x+5 | D. | y=$\frac{2}{3}x-\frac{1}{3}$ |
4.若3$<(\frac{1}{3})$x<27,则( )
A. | -1<<3 | B. | -3<<-1 | C. | x<-1或x>3 | D. | 1<x<3 |