Calcul de limites

Indéterminations [(k/0)] et [(0/0)]

n=0;
n++;
f[n]:=Expand[(x+3)*(2x-1)]/(3-x);
lim[n]=Limite[f[n],x,3];

n++;
f[n]:=(x^2-3x-4)/(x^2-16);
lim[n]=Limite[f[n],x,4];

n++;
f[n]:=Expand[(2-x)*(2x+1)]/Expand[(2x+1)*(x+1)];
lim[n]=Limite[f[n],x,-1/2];

n++;
f[n]:=(5-3x)/Expand[(2-x)(3+x)];
lim[n]=Limite[f[n],x,2];

n++;
f[n]:=Expand[(x-4)*(x+2)]/Expand[(x+2)*(3x-4)];
lim[n]=Limite[f[n],x,-2];

n++;
f[n]:=(x^2-3)/(x^2-3x);
lim[n]=Limite[f[n],x,0];

n++;
f[n]:=Expand[(x)*(x-3)*(x+1)]/Expand[(x+1)*(x-2)];
lim[n]=Limite[f[n],x,-1];

n++;
f[n]:=Expand[(x-1)*(3x+2)]/Expand[(x-1)*(x+1)*(x+2)];
lim[n]=Limite[f[n],x,1];

n++;
f[n]:=Expand[(x-1)*(2x-3)*(x+2)]/(1-x)^2;
lim[n]=Limite[f[n],x,1];

n++;
f[n]:=Expand[(3-5x)*(x+2)]/Expand[(5x-3)x];
lim[n]=Limite[f[n],x,3/5];

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

Solutions

Do[Print[i,") "];Print[Detaille[lim[i]]],{i,1,n}]