洛必达求极限 limsinxlnx x趋近于0+,lim(2/π·arctanx)^x x趋近无穷大,lim(ln1(/

洛必达求极限 limsinxlnx x趋近于0+,lim(2/π·arctanx)^x x趋近无穷大,lim(ln1(/x))^x x趋近0+,
limlnx·ln(1+x) x趋近0+,lim(x^3+x^+x+1)^1/3-x x趋近无穷大,lim (e^x-e^sinx)/(x-sinx) x趋近0,lim(sinx/x)^(1/x^2)x趋近0,lim[1/e· (1/+x)^(1/x)]^(1/x) x趋近0

1.lnx*ln(1+x)=ln(1+x)/(1/lnx)
=[1/(1+x)][-1/(lnx)^2*1/x]
=x(lnx)^2/(1+x)
=(lnx)^2/(1+1/x)
=[2lnx/x]/(-1/x^2)
=-2lnx/(1/x)
=(-2/x)/(-1/x^2)
=2x=0
2.(x^3+x^+x+1)^1/3-x
=[(x^3+x^+x+1)^1/3-x]
*[(x^3+x^+x+1)^2/3+(x^3+x^+x+1)^1/3*x+x^2]/[(x^3+x^+x+1)^2/3+(x^3+x^+x+1)^1/3*x+x^2]
=[x^3+x^2+x+1-x^3]/[(x^3+x^+x+1)^2/3+(x^3+x^+x+1)^1/3*x+x^2]
=(x^2+x+1)/[(x^3+x^+x+1)^2/3+(x^3+x^+x+1)^1/3*x+x^2]
上下同除x^2
=(1+1/x+1/x^2)/[(1+1/x+1/x^2+1/x^3)^(2/3)+(1+1/x+1/x^2+1/x^3)^(1/3)+1]
=(1+0+0)/(1+1+1)=1/3
3.(e^x-e^sinx)/(x-sinx)
=(e^x-e^sinx *cosx)/(1-cosx)
=(e^x-e^sinx *(cosx)^2+e^sinx*cosx*sinx)/sinx
=(e^x-e^sinx *(cosx)^3+2e^sinx *cosx sinx+e^sinx*(cosx)^2*sinx+e^sinx*(cos 2x))/cosx
=(1-1+0+0+1)/1=1
4.令y=(sinx/x)^(1/x^2)
lny=(ln sinx -ln x)/x^2
=(cosx/sinx -1/x)/2x
=(-csc^2 x +1/x^2)/2
=(1/x^2-1/sinx^2)/2
=[(sinx)^2-x^2]/(2x^2 sinx^2)
=[2sinx cosx -2x]/[4x sinx^2+4x^2sinx cosx]
=[sin2x-2x]/[2x(1-cos 2x)+2x^2 sin 2x]
=[2cos 2x -2]/[2(1-cos 2x)+4x sin 2x +4x sin2x +4x^2cos 2x]
=[cos2x-1]/[1-cos2x+4xsin2x+2x^2cos2x]
=[-2sin2x]/[2sin2x+4sin2x+8xcos2x+4xcos2x-4x^2sin2x]
=-4cos2x/[4cos2x+8cos2x+8cos2x-16xsin2x+4cos2x-8xsin2x-8xsin2x-8x^2cos2x]
=-4/[4+8+8+0+4-0-0-0]
=-4/24=-1/6
y->exp(-1/6)
5.令y=[1/e· (1/+x)^(1/x)]^(1/x)
lny=ln[1/e· (1/+x)^(1/x)]/x
=[(ln(1+x))/x-1]/x
=[ln(1+x)-x]/x^2
=[1/(1+x)-1]/2x
=[-1/(1+x)^2]/2
=-1/2
y->e^(-1/2)