If [ ] denotes the greatest integer function then f(x)=[x]+[x+12]
f(x)=[x]+[x+12]f(x)x→1−/2=limh→0[12−h]+[12−h+12]f(x)x→1−/2=limh→00+[1−h]=limh→00+1+[−h]=1−1=0f(x)x→1+/2=limh→0[12+h]+[12+h+12]f(x)x→1+/2=0+1=1f(x)x→1−/2≠f(x)x→1+/2
So, function is discontinuous at x=12.