Exercices: Calcul de limites vers l'infini

n=0;

n++;
l[n]=Limite[(x^2-3x+5)/(3x^2-x+3),x,-∞];

n++;
l[n]=Limite[(2-3x)/(5x^2-3),x,+∞];

n++;
l[n]=Limite[(2x-3)/(4-3x),x,+∞];

n++;
l[n]=Limite[(5x^2-x+3)/(3+4x),x,-∞];

n++;
l[n]=Limite[(Sqrt[4x^2-3x+5]),x,-∞];

n++;
l[n]=Limite[2x+(Sqrt[x-1]),x,-∞];

n++;
l[n]=Limite[2x+(Sqrt[x-1]),x,+∞];

n++;
l[n]=Limite[(3x-1)/(Sqrt[x^2+x+2]),x,+∞];

n++;
l[n]=Limite[(3x-1)/(Sqrt[x^2+x+2]),x,-∞];

n++;
l[n]=Limite[((Sqrt[2x-1])/(x+1)),x,+∞];

n++;
l[n]=Limite[(3x-1)/(Sqrt[3-x]),x,-∞];

n++;
l[n]=Limite[(Sqrt[2x^2-x+6])+x,x,-∞];

n++;
l[n]=Limite[(Sqrt[x^2-3x+1])-x,x,+∞];

n++;
l[n]=Limite[(Sqrt[x^2-3x+1])-x,x,-∞];

n++;
l[n]=Limite[(Sqrt[4x^2-3x+2])+2x,x,-∞];

n++;
l[n]=Limite[(Sqrt[4x^2-3x+2])+2x,x,+∞];



Do[Print[i,") ",l[i]],{i,1,n}]

Do[Print[i,") ",Presente[l[i]]],{i,1,n}]