Last updated at March 10, 2021 by

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Ex 5.1, 5 Is the function f defined by π(π₯)={β(π₯, ππ π₯β€1@&5, ππ π₯>1)β€ continuous at π₯ = 0 ? At π₯ = 1 ? At π₯ = 2 ? Given π(π₯)={β(π₯, ππ π₯β€1@&5, ππ π₯>1)β€ At x = 0 For x = 0, f(x) = x Since this a polynomial It is continuous β΄ f(x) is continuous for x = 0 At 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) π(π₯)={β(π₯, ππ π₯β€1@&5, ππ π₯>1)β€ 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) = 1 β 0 = 1 RHL at x β 1 limβ¬(xβ1^+ ) f(x) = limβ¬(hβ0) f(1 + h) = limβ¬(hβ0) 5 = 5 Since L.H.L β R.H.L f(x) is not continuous at x = 1 At x = 2 For x = 2, f(x) = 5 Since this a constant function It is continuous β΄ f(x) is continuous for x = 2

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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.