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Chapter 13 Class 11 Limits and Derivatives (Term 1 and Term 2)

Serial order wise

Last updated at Nov. 30, 2019 by Teachoo

Ex 13.1, 30 If f(x) = {β(|x|+1, x<0@0 x=0@|x| β1, x>0)β€ . For what value (s) of a does (πππ)β¬(π₯βπ) f(x) exists? We need to find value of a for which limβ¬(xβa) f(x) exists We check limit different values of a When a = 0 When a < 0 When a > 0 Case 1: When a = 0 Limit exists at a = 0 if limβ¬(xβ0^+ ) " f(x) = " limβ¬(xβ0^β ) " f(x)" f(x) = {β(|x|+1, x<0@0 x=0@|x| β1, x>0)β€ . LHL at x β 0 limβ¬(xβ0^β ) f(x) = limβ¬(hβ0) f(0 β h) = limβ¬(hβ0) f(β h) = limβ¬(hβ0) |ββ| + 1 = limβ¬(hβ0) β + 1 = 0 + 1 = 1 RHL at x β 0 limβ¬(xβ0^+ ) f(x) = limβ¬(hβ0) f(0 + h) = limβ¬(hβ0) f(h) = limβ¬(hβ0) |β| β 1 = limβ¬(hβ0) β β 1 = 0 β 1 = β1 Since 1 β β 1 So, left hand limit and right hand limit are not equal at x = 0 Hence, limβ¬(xβ0) f(x) does not exist β΄ At x = 0, Limit does not exist Case 2: When a < 0 For a < 0 f(x) = |π₯|+1 f(x) = βπ₯+1 Since this a polynomial It is continuous β΄ Limit exists for a < 0 (As x is negative) Case 3: When a > 0 For a < 0 f(x) = |π₯|β1 f(x) = π₯+1 Since this a polynomial It is continuous β΄ Limit exists for a > 0 Therefore, we can say that (πππ)β¬(π₯βπ) f(x) exists for all a, where a β 0 (As x is positive)