lim sin[π(n^2+1)^1/2] n趋向无穷求极限 求详解

lim sin[π(n^2+1)^1/2] n趋向无穷求极限 求详解

lim sin[π(n^2+1)^1/2]=lim {sin[π(n^2+1)^1/2]-sin nπ}
=lim cos(π/2)((n^2+1)^1/2+n) sin (π/2)[(n^2+1)^1/2-n)]
=lim cos(π/2)((n^2+1)^1/2+n) sin (π/2){1/[(n^2+1)^1/2+n)]}.
cos(π/2)((n^2+1)^1/2+n)有界，lim sin (π/2){1/[(n^2+1)^1/2+n)]}=0
有界与无穷小之积仍是无穷小。所以
.lim sin[π(n^2+1)^1/2]=0

lim sin[π(n^2+1)^(1/2)]
=lim (-1)^n*sin[π(n^2+1)^(1/2)-nπ]
=lim (-1)^n*sin[π/【(n^2+1)^(1/2)+n】]
=0。最后一个等式是无穷小量乘以有界量

求极限 n➔∞lim sin[π√(n²+1)]
解：当n=1时，sin(π√2)=sin(1.4142....π)=sin(π+0.4142....π)=-sin(0.4142....π)
当n=2时，sin(π√5)=sin(2.2360.....π)=sin(2π+0.2360...π)=sin(0.2360.....π)
当n=3时，sin(π√10)=sin(3.1622...π)=sin(3π+0.1622...π)=-sin(0.1622...π)
当n=4时，sin(π√17)=sin4.12310...π)=sin(4π+0.12310...π)=sin(0.12310...π)
当n=5时，sin(π√26)=sin(5.0990...π)=sin(5π+0.0990...π)=-sin(0.0990.....π)
当n=6时，sin(π√37)=sin(6.08276...π)=sin(6π+0.08276...π)=sin(0.08276...π)
................................................................................................................
可以预期，n➔∞lim sin[π√(n²+1)] ＝0
理论上的严格证明，似乎有点难。