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

Transcript

Ex 13.1, 24 (Method 1) Find (π‘™π‘–π‘š)┬(π‘₯β†’1) f(x), where f(x) = {β–ˆ(π‘₯2 βˆ’1,@βˆ’π‘₯2 βˆ’1,)─ β– 8(π‘₯ ≀1@π‘₯>1) The Limit at x = 1 will be (π‘™π‘–π‘š)┬(π‘₯β†’1) f(x) = lim┬(γ€–xβ†’1γ€—^βˆ’ ) f(x) =(π‘™π‘–π‘š)┬(γ€–π‘₯β†’1γ€—^+ ) f(x) (π’π’Šπ’Ž)┬(γ€–π’™β†’πŸγ€—^βˆ’ ) f(x) = (π‘™π‘–π‘š)┬(π‘₯β†’1) x2 – 1 = (1)2 – 1 = 1 – 1 = 0 (π’π’Šπ’Ž)┬(γ€–π’™β†’πŸγ€—^+ ) f(x) = (π‘™π‘–π‘š)┬(π‘₯β†’1) (–x2 – 1) = –(1)2 – 1 = –1 – 1 = –2 Thus, (π’π’Šπ’Ž)┬(γ€–π’™β†’πŸγ€—^+ )f(x) β‰  (π’π’Šπ’Ž)┬(γ€–π’™β†’πŸγ€—^βˆ’ )f(x) Since, Left Hand Limit & Right Hand Limit are not equal Hence (π₯π’Šπ’Ž)┬(π’™β†’πŸ) f(x) does not exit Ex 13.1, 24 (Method 2) Find (π‘™π‘–π‘š)┬(π‘₯β†’1) f(x), where f(x) = {β–ˆ(π‘₯2 βˆ’1,@βˆ’π‘₯2 βˆ’1,)─ β– 8(π‘₯ ≀1@π‘₯>1) The Limit at x = 1 will be (π‘™π‘–π‘š)┬(π‘₯β†’1) f(x) = lim┬(γ€–xβ†’1γ€—^βˆ’ ) f(x) =(π‘™π‘–π‘š)┬(γ€–π‘₯β†’1γ€—^+ ) f(x) LHL at x β†’ 1 lim┬(xβ†’1^βˆ’ ) f(x) = lim┬(hβ†’0) f(1 βˆ’ h) = lim┬(hβ†’0) (1 βˆ’ h)2 βˆ’1 = (1 βˆ’ 0)2 βˆ’ 1 = (1)2 βˆ’ 1 = 1 βˆ’ 1 = 0 RHL at x β†’ 1 lim┬(xβ†’1^+ ) f(x) = lim┬(hβ†’0) f(1 + h) = lim┬(hβ†’0) βˆ’(1 + h)2 βˆ’ 1 = βˆ’(1 + 0)2 βˆ’ 1 = βˆ’(1)2 βˆ’ 1 = βˆ’1 βˆ’ 1 = βˆ’2 Since LHL β‰  RHL ∴ (π’π’Šπ’Ž)┬(π’™β†’πŸ) f(x) doesn’t exist

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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.