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Ex 12.1, 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 12.1, 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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo