bonsoir j ai besoi d'aide slvp [tex]f(x) = \sqrt{ \frac{x {}^{3} }{x + 2} } [/tex] calculer lla limite lim ( f(x)-x ) quand x tend vers [tex] + \infty [/tex]

f(x)=√(x³/(x+2))
donc f(x)-x=√(x³/(x+2))-x=(√(x³/(x+2))-x)(√(x³/(x+2))+x)/(√(x³/(x+2))+x)
                  =(x³/(x+2)-x²)/(√(x³/(x+2))+x)
                  =(x³-(x+2)x²)/((x+2)(√(x³/(x+2))+x)
                  =(-2x²)/((x+2)(√(x³/(x+2))+x)
                  =-2/((1+2/x)(√(x/(x+2))+1)

or lim((√(x/(x+2))+1),x→+∞)=1
lim(1+2/x,x→+∞)=1
lim(-2,x→+∞)=-2
par opérations on déduit que : lim(f(x)-x,x→+∞)=-1