Plz explain example 5 Share with your friends Share 0 Gursheen Kaur answered this Dear Student, Solution) fx=x+1, x<0 =0, x=0 =x-1, x>0Function definiton changes at x=0L.H.L.=limx→0- fx=limx→0- x+1=0+1L.H.L.=1 R.H.L.=limx→0+ fx=limx→0+ x-1=0-1R.H.L.=-1As L.H.L.≠R.H.L., hence limit does not exist at x=0 but it will exist at every otherpoint.Hence a∈R-0 Regards! 0 View Full Answer