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Ex 13.1, 30 - If f(x) = { |x|+1, 0, |x|-1. For what values of a, does

Ex 13.1, 30 - Chapter 13 Class 11 Limits and Derivatives - Part 2
Ex 13.1, 30 - Chapter 13 Class 11 Limits and Derivatives - Part 3 Ex 13.1, 30 - Chapter 13 Class 11 Limits and Derivatives - Part 4

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Ex 13.1, 30 If f(x) = {β–ˆ(|x|+1, x<[email protected] [email protected]|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<[email protected] [email protected]|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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.