limite Log

calculer la limite suivante :

Bonjour;

On a : 1/(x-1)-1/Ln(x)=1/y-1/Ln(1+y)=(Ln(1+y)-y)/(yLn(1+y)) , avec x=1+y .

Et comme Ln(1+y)=y-(y²)/2+(y^3)/3-(y^4)/4+o(y^4) alors :

Ln(1+y)-y=y-(y²)/2+(y^3)/3-(y^4)/4-y+o(y^4)=-(y²)/2+(y^3)/3-(y^4)/4+o(y^4)

et yLn(1+y)=y²-(y^3)/2+(y^4)/3+o(y^4) ,

donc Ln(1+y)-y=y²(-1/2+y/3-(y²)/4+o(y²)) et yLn(1+y)=y²(1-y/2+(y²)/3+o(y²))

donc (Ln(1+y)-y)/(yLn(1+y))=(-1/2+y/3-(y²)/4+o(y²))/(1-y/2+(y²)/3+o(y²))=-1/2+y/12+-(y²)/24+o(y²) ,

donc lim(x-->1+) 1/(x-1)-1/Ln(x) = lim(y-->0+)(Ln(1+y)-y)/(yLn(1+y))=lim(y-->0+)-1/2+y/12+-(y²)/24+o(y²)=-1/2 .