Let f(x)=[x]  and g(x)={0,x∈Ix2,x∉I  then

Ltx→1g(x)=Ltx→1x2=12=1  and g(1)=0 Ltx→1−0f(x)=Ltx→1−0[x]=0 Ltx→1+0f(x)=Ltx→1+0[x]=1 (gof)(x)=g(f(x))=g([x])=0                                       [∵ [x] is an integer] Hence (gof)(x)=0= constant(fog)(x)=f[(g)(x)]={f(0), x∈If(x2),x∉I ={[0],x∈I[x2],x∉I ={0,x∈I[x2],x∉I Clearly (fog)(x)  is continuous at x=0  and is continuous at all other integral points.