極限問題 無限.三角

1
lim(x->∞)_(1－1/4)(1－1/9)....(1－1/n^2)
=lim(x->∞)_[(1－1/2)*(1+1/2)]*[(1－1/3)*(1+1/3)…..*[(1－1/n)*(1+1/n)]
=lim(x->∞)_[(1－1/2)*(1+1/2)]*[(1－1/3)*(1+1/3)]*[(1－1/4)*(1+1/4)]
*[(1-1/5)*(1+1/6)]*…..*[(1－1/n)*(1+1/n)]
=lim(x->∞)_[(1/2)*(3/2)]*[(3/2)*(4/3)]*[(3/4)*(5/4)]*[(4/5)*(6/5)]
*…*[(n－1)/n*(n+1)/n]
=lim(x->∞)_(1/2)*[(n+1)/n]
=1/2

2
lim(x->∞)_[x^(1/3)*(x+2)^(2/3)－x]
Sol
Set a=x^(1/3)*(x+2)^(2/3)，b=x
a^2=x^(2/3)*(x^2+4x+4)^(2/3)=(x^3+4x^2+4x)^(2/3)
ab=x*(x^1/3)*(x+2)^(2/3)=(x^2)^(2/3)*(x+2)^(2/3)=(x^3+2x^2)^(2/3)
a^3=x(x+2)^2=x^3+4x^2+4x
a^2+ab+b^2=(x^3+4x^2+4x)^(2/3)+x^(4/3)*(x+2)^(2/3)+x^2
a^3－b^3=x^3+4x^2+4x－x^3=4x^2+4x
A=lim(x->∞)_[x^(1/3)*(x+2)^(2/3)－x]
=lim(x->∞)_(a－b)
=lim(x->∞)_(a^3－b^3)/(a^2+ab+b^2)
=lim(x->∞)_(4x^2+4x)/[(x^3+4x^2+4x)^(2/3)+x^(4/3)*(x+2)^(2/3)+x^2]
=lim(x->∞)_(4+4/x)/[(1+4/x+4/x^2)^(2/3)+1*(1+2/x)^(2/3)+1]
(4+0)/[(1+0+0)^(2/3)+1*(1+0)^(2/3)+1]
=4/3
or
lim(x->∞)_[x^(1/3)*(x+2)^(2/3)－x]
=lim(x->∞)_[x^(-2/3)*(x+2)^(2/3)－1]/(1/x)
=lim(x->∞)_[(1+2/x)^(2/3)－1]/(1/x)
=lim(y->0+)_[(1+2y)^(2/3)－1]/y 0/0 type
=lim(y->0+)_2*(2/3)*(1+y)^(－1/3)
=4/3

3
lim(x->0)_(Cos2x－4Cosx+3)/x^4
=lim(x->0)_(2Cos^2 x－4Cosx+2)/x^4
=2*lim(x->0)_(Cos^2 x－2Cosx+1)/x^4
=2*lim(x->0)_(Cosx－1)^2/x^4
=2*lim(y->0)_(Cos2y－1)^2/(16y^4)
=(1/8)*lim(y->0)_(2Sin^2 y)^2/y^4
=(1/2)*lim(y->0)_Sin^4 y/y^4
=1/2
or
lim(x->0)_(Cos2x－4Cosx+3)/x^4 0/0 type
=lim(x->0)_(－2Sin2x+4Sinx)/(4x^3) 0/0 type
=lim(x->0)_(－4Cos2x+4Cosx)/(12x^2) 0/0 type
=lim(x->0)_(8Sin2x－4Sinx)/(24x) 0/0 type
=lim(x->0)_(16Cos2x－4Cosx)/24
=(16－4)/24
=1/2

1.
原式
= (1+1/2)(1-1/2)(1+1/3)(1-1/3).....(1+1/n)(1-1/n)
= [(1+1/2)(1+1/3).....(1+1/n)] [(1-1/2)(1-1/3).....(1-1/n)]
= [(3/2)(4/3).....(n+1)/n] [(1/2)(2/3).....(n-1)/n]
= [ (n+1)/2 ] [1/n]
= (1/2)(1+1/n) → 1/2 when n→∞
Ans: 1/2