lim (cos1/n)^n^2 =lim e^[n^2*ln(cos1/n)]=e^{lim ln(cos1/n)/(1/n^2) ("0/0"罗比塔法则)
n->∞ n->∞ n->∞
=e^{lim 1/(cos1/n)*(-sin1/n)*(-1/n^2)/(-2/n^3)=e^{lim (-sin1/n)*n/2}=e^{lim (sin1/n)/(1/n)*(-1/2)}
n->∞ n->∞ n->∞
=e^(-1/2)
补充题:
lim (1+|x|)^1/x
x->0
x->+0 lim (1+|x|)^1/x=lim (1+x)^1/x=e
x->-0 lim (1+|x|)^1/x=lim (1-x)^1/x=lim [(1-x)^(-1/x)]^(-1)=e^(-1)=1/e
故lim (1+|x|)^1/x x->0极限不存在
cos(1/n)^n^2={[1+(cos1/n-1)]^[1/(cos1/n-1)]}^[n^2*(cos1/n-1)]
={[1+(cos1/n-1)]^[1/(cos1/n-1)]}^[n^2*(sin1/n)^2/(cos1/n+1)]
limn->∞n^2*(sin1/n)^2=limn->∞[(sin1/n)/n]^2=1
limn->∞(cos1/n+1)=2
limn->∞{[1+(cos1/n-1)]^[1/(cos1/n-1)]}=e
所以,limn->∞(cos1/n)^n^2=√e
cos2X=1-2*(sinX)^2
lim(cos1/n)^n^2
=lim[(1-2sin^2(1/2n))^(-1/2sin^2(1/2n))]^[n^2*2sin^2(1/2n))]
=e^lim[n^2*2sin^2(1/2n))]
=e^(1/2)
lim (1+|x|)^1/x x->0
x>0且 x->0 lim (1+|x|)^1/x=e
x<0且 x->0 lim (1+|x|)^1/x=e^(-1)=1/e
故lim (1+|x|)^1/x x->0极限不存在