Question

为什么这个等于0而不是1/2？

Answer 1

lim(n->∞) [ n.sin(1/n)- cos(1/n) ] /ln(1+1/n^2)

consider
y->0+
siny = y- (1/6)y^3 +o(y^3)
(1/y)siny = 1- (1/6)y^2 +o(y^2)
cosy = 1 -(1/2)y^2 +o(y^2)
(1/y)siny - cosy = (1/3)y^2 +o(y^2)
lim(x->∞) [ x.sin(1/x)- cos(1/x) ] /ln(1+1/x^2)
y=1/x
=lim(y->0+) [ (1/y).siny- cosy ] /ln(1+y^2)
=lim(y->0+) [ (1/y).siny- cosy ] / y^2
=lim(y->0+) (1/3)y^2 / y^2
=1/3
=>
lim(n->∞) [ n.sin(1/n)- cos(1/n) ] /ln(1+1/n^2) =1/3

Answer 2

令u=1/n趋于零
lim=lim(sinu－ucosu)/uln(1+u²)
=lim(sinu－ucosu)/u³
=lim(cosu－cosu+usinu)/3u²
=1/3

Answer 3

先看分母：ln(1+1/n²）~1/n²
sin(1/n)~1/n-1/(6n³)
nsin(1/n)~1-1/(6n²)
cos(1/n)~1-1/(2n²)
nsin(1/n)-cos(1/n)~1-1/(6n²)-[1-1/(2n²)]=1/(3n²)
因此结果是1/3。
不是0，也不是1/2