Lim(x趋向无穷)[(x^2-6x+8)/(x^2-5x+4)]?怎么做呀?
Lim(x趋向无穷)[(x^2-6x+8)/(x^2-5x+4)]?????????怎么做呀?
lim(x趋向无穷) (x^2-6x+8)/ (x^2-5x+4)
=lim(x趋向无穷) (x-2)(x-4)/ ( (x-1)(x-4) )
=lim(x趋向无穷) (x-2)/(x-1)
=lim(x趋向无穷)(x-1)-1/(x-1)
=lim(x趋向无穷){1-1/(x-1)}
x趋向无穷时极限等于1
lim x(arctg((x+1)/(x+2))-∏/4),x趋向无穷
极限转化为lim[(arctg((x+1)/(x+2))-∏/4)/(1/x)]，为0比0型极限
用洛必达法则，对分子分母同时求导，极限化为
lim[-x^2/(2x^2+6x+5)]
=lim[-1/(2+6/x+5/x^2)]
6/x,5/x^2都是0，所以极限=-1/2
lim x趋向无穷x+2cosx/x的平方+6
(x-2)/(x^2+6)≤(x+2cosx)/(x^2+6)≤(x-2)/(x^2+6)
lim(x->∞)(x-2)/(x^2+6)≤lim(x->∞)(x+2cosx)/(x^2+6) ≤lim(x->∞)(x-2)/(x^2+6)
0≤lim(x->∞)(x+2cosx)/(x^2+6)≤0
=>
lim(x->∞)(x+2cosx)/(x^2+6) =0
lim〈x趋向无穷〉 ln（1+2x）/x
lim(x->inf) ln(1+2x) / x
=lim ln[(1+2x)^(1/x)]
=ln lim [(1+2x)^(1/2x * 2)]
=ln (1)^lim 2，极限lim(x->inf) (1+x)^(1/x) = 1
=ln (1)
=0
lim x趋向无穷大 (1+2/x)^x=?
lim x 趋近无穷大 则2/x趋近0 所以等于1
lim(x趋向无穷）（（3-2x）/（2-2x））^x
(3-2x)/(2-2x)=1+1/(2-2x)
令t=1/（2-2x），则x=1-（1/2t）则x→无穷时，t→0
lim(x趋向无穷）（（3-2x）/（2-2x））^x
=lim（t→0）（1+t）^[1-（1/2t）]
=lim（t→0）（1+t）/√[（1+t）^1/t]
=1/√e
=e^-1/2
lim((arctanx+π)/(x+2)) x趋向无穷
x→-∞，arctanx→-π/2
lim(x→-∞)((arctanx+π)/(x+2)) =lim(x→-∞)(π/2)/[x+2]=0
x→-∞，arctanx→π/2
lim(x→+∞)((arctanx+π)/(x+2)) =lim(x→+∞)(3π/2)/[x+2]=0
∵lim(x→-∞)f(x)=lim(x→+∞)f(x)=0
∴linm(x→∞）f(x)=0
题，求， lim(x趋向无穷) x^3+4x^2-5 / 2x^3-2x+1 =?
lim(x→∞)(x^3+4x^2-5)/(2x^3-2x+1)=lim(x→∞)(x^3-x+0.5)/(2x^3-2x+1)+lim(x→∞)(4x^2+x-5.5)/(2x^3-2x+1)=1/2+lim(x→∞)(4+1/x-5.5/x^2)/(2x-2/x+1/x^2)=1/2
lim x趋向无穷大(1-2/x)^2=？
1
求采纳
lim（x/（1+x^2））,x趋向无穷大
当X趋于无穷大；
lim[x/(1+x^2)]
=lim(1/x)/(1/x^2 +1)
=0/(0+1)
=0
