Last updated at Aug. 13, 2021 by Teachoo

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Ex 5.1, 14 Discuss the continuity of the function f, where f is defined by π(π₯)={β(3, ππ 0β€π₯β€1@4, ππ 1<π₯<3@ 5, ππ 3β€π₯β€10)β€ Since we need to find continuity at of the function We check continuity for different values of x When 0 β€ x < 1 When x = 1 When 1 < x < 3 When x = 3 When 3 < x β€ 10 Case 1 : When 0 β€ x < 1 For 0 β€ x < 1, f(x) = 3 Since this constant It is continuous β΄ f(x) is continuous for 0 β€ x < 1 Case 2 : 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) 3 = 3 RHL at x β 1 limβ¬(xβ1^+ ) f(x) = limβ¬(hβ0) f(1 + h) = limβ¬(hβ0) 4 = 4 Since L.H.L β R.H.L f(x) is not continuous at x = 1 Case 3 : When 1 < x < 3 For 1 < x < 3 f(x) = 4 Since this constant It is continuous β΄ f(x) is continuous for 1 < x < 3 Case 4 : When x = 3 f(x) is continuous at π₯ =3 if L.H.L = R.H.L = π(3) if limβ¬(xβ3^β ) π(π₯)=limβ¬(xβ3^+ ) " " π(π₯)= π(3) Since L.H.L β R.H.L f(x) is not continuous at x=3 Since there are two different functions on the left & right of 3, we take LHL & RHL . LHL at x β 3 limβ¬(xβ3^β ) f(x) = limβ¬(hβ0) f(3 β h) = limβ¬(hβ0) 4 = 4 RHL at x β 3 limβ¬(xβ3^+ ) f(x) = limβ¬(hβ0) f(3 + h) = limβ¬(hβ0) 5 = 5 Case 4 : When x = 3 f(x) is continuous at π₯ =3 if L.H.L = R.H.L = π(3) if limβ¬(xβ3^β ) π(π₯)=limβ¬(xβ3^+ ) " " π(π₯)= π(3) Since L.H.L β R.H.L f(x) is not continuous at x=3 Case 5: When 3 < x β€ 10 For 3 < x β€ 10, f(x) = 5 Since this constant It is continuous β΄ f(x) is continuous for 3 < x β€ 10 Hence points of discontinuity are x = 1 & x = 3 Thus, f is continuous for 0 β€ x β€ 10 except x β {π,π}

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About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.