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Last updated at April 16, 2024 by Teachoo

Ex 5.1, 13 Is the function defined by π(π₯)={ β(π₯+5, ππ π₯β€1@& π₯β5 , ππ π₯>1)β€ a continuous function? Since we need to find continuity at of the function We check continuity for different values of x When x = 1 When x < 1 When x > 1 Case 1 : When x = 1 f(x) is continuous at π₯ =1 if L.H.L = R.H.L = π(1) if limβ¬(xβ1^β ) π(π₯)=limβ¬(xβ1^+ ) " " π(π₯)= π(1) Since there are two different functions on the left & right of 1, we take LHL & RHL . LHL at x β 1 limβ¬(xβ1^β ) f(x) = limβ¬(hβ0) f(1 β h) = limβ¬(hβ0) ((1ββ)+5) = (1β0)+5 = 1 + 5 = 6 RHL at x β 1 limβ¬(xβ1^+ ) f(x) = limβ¬(hβ0) f(1 + h) = limβ¬(hβ0) ((1+β)β5) = (1+0)β5 = 1 β 5 = β4 Since L.H.L β R.H.L β΄ f is not continuous at x = 1 Case 2 : When x < 1 For x < 1, f(x) = x + 5 Since this a polynomial It is continuous β΄ f(x) is continuous for x < 1 Case 3 : When x > 1 For x > 1, f(x) = x β 5 Since this a polynomial It is continuous β΄ f(x) is continuous for x > 1 Hence, only π₯=1 is point of discontinuity. β΄ f is continuous at all real numbers except 1 Thus, f is continuous for πβ R β {1}