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  1. Chapter 13 Class 11 Limits and Derivatives
  2. Serial order wise

Transcript

Ex 13.1, 30 If f(x) = {β–ˆ(|x|+1, x<0@0 x=0@|x| βˆ’1, x>0)─ . For what value (s) of a does (π‘™π‘–π‘š)┬(π‘₯β†’π‘Ž) f(x) exists? We need to find value of a for which lim┬(xβ†’a) 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 lim┬(xβ†’0^+ ) " f(x) = " lim┬(xβ†’0^βˆ’ ) " f(x)" f(x) = {β–ˆ(|x|+1, x<0@0 x=0@|x| βˆ’1, x>0)─ . LHL at x β†’ 0 lim┬(xβ†’0^βˆ’ ) f(x) = lim┬(hβ†’0) f(0 βˆ’ h) = lim┬(hβ†’0) f(βˆ’ h) = lim┬(hβ†’0) |βˆ’β„Ž| + 1 = lim┬(hβ†’0) β„Ž + 1 = 0 + 1 = 1 RHL at x β†’ 0 lim┬(xβ†’0^+ ) f(x) = lim┬(hβ†’0) f(0 + h) = lim┬(hβ†’0) f(h) = lim┬(hβ†’0) |β„Ž| – 1 = lim┬(hβ†’0) β„Ž – 1 = 0 – 1 = –1 Since 1 β‰  – 1 So, left hand limit and right hand limit are not equal at x = 0 Hence, lim┬(xβ†’0) f(x) does not exist ∴ At x = 0, Limit does not exist Case 2: When a < 0 For a < 0 f(x) = |π‘₯|+1 f(x) = βˆ’π‘₯+1 Since this a polynomial It is continuous ∴ Limit exists for a < 0 (As x is negative) Case 3: When a > 0 For a < 0 f(x) = |π‘₯|βˆ’1 f(x) = π‘₯+1 Since this a polynomial It is continuous ∴ Limit exists for a > 0 Therefore, we can say that (π‘™π‘–π‘š)┬(π‘₯β†’π‘Ž) f(x) exists for all a, where a β‰  0 (As x is positive)

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.