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Let ` f(x) = (x^(2) -1)/(x-1) ` <br> Now, ` f(1)= 0/0` , which is meaningless. <br> So, the funcation it not defined at x=1. <br> Also, when ` x ne 1`, we have <br> ` f (x) = (( x^(2)-1)/(x-1))= (x +1)` <br> The being a linear funcation, its graph is a straight line. <br> Some, of the points on the graph are (0,1) , (-1,0) , (2,3) ,(3,4) (-2, -=1) , (-3,-2) etc. <br> Joining these points, we obtain the required graph. Clearly, the point (1,2) does not lie on the graph. so, it is a broken graph , and we shall say that given function is discontinuous at x= 1. <br> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/RSA_MATH_XII_C09_S01_008_S01.png" width="80%">