LIMITES

BONSOIR!!!
Je vs propose lé lim suivante
1) lim(x-->+00)x(Arctan{x/(x+1) -pi/4}
2)lim(x-->0+){(3°V(1+3°Vx^2)-4°V(1-3°Vx^2)}/Vx
3) lim(x-->+00)(5°V(x^2+1)-(3°V(x^2+x+1)

NB (3°Vx) veu dire racine cubik de x

1ere limite pose : t= Arctan{x/(x+1) -pi/4 quands x tends vers + OO

t tends vers 0 puis tu cherche x

é pr la lim(x-->+00)(5°V(x^2+1)-(3°V(x^2+x+1)

(x)^(2/5) *(1+(1/x²))^(1/5)-(x)^(2/3)(1+1/x+1/x²)^(1/3)
=
x^(2/3)*(x^(2/5-2/3)*(1+1/x²)^(1/5)-(1+1+1/x²)^(1/3))
2/5-2/3<0 donc lim+00x^(2/5-2/3)=0
=>lim+00x^(2/5-2/3)*(1+1/x²)^(1/5)-(1+1+1/x²)^(1/3)=-1
=+>lim+005°V(x^2+1)-(3°V(x^2+x+1)=-00
sauf erreur

mais moi j compen ps bien les sign si c possibl ecrivent bien plzz