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Ex 13.1, 31 - Chapter 13 Class 11 Limits and Derivatives (Term 1 and Term 2)

Last updated at Sept. 6, 2021 by

Ex 13.1, 31 - If function f(x) lim x->1 f(x)-2/x2-1 = pi

Ex 13.1, 31 - Chapter 13 Class 11 Limits and Derivatives - Part 2

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Ex 13.1, 31 If the function f(x) satisfies lim┬(x β†’ 1) (𝑓(π‘₯) βˆ’ 2)/(π‘₯2 βˆ’ 1) = Ο€ , evaluate lim┬(xβ†’1) f(x) . Given lim┬(xβ†’1) (𝑓(π‘₯) βˆ’ 2)/(π‘₯^2 βˆ’ 1) = Ο€ (lim┬(xβ†’1) 𝑓(π‘₯) βˆ’ 2)/(lim┬(xβ†’1) γ€–(π‘₯γ€—^2 βˆ’ 1) ) = Ο€ lim┬(xβ†’1) (f(x) – 2) = Ο€ Γ— lim┬(xβ†’1) (x2 – 1) lim┬(xβ†’1) f(x) – lim┬(xβ†’1) 2 = Ο€ (lim┬(xβ†’1) x2 – lim┬(xβ†’1) 1) By Algebra of limits (π‘™π‘–π‘š)┬(π‘₯β†’π‘Ž) (𝑓(π‘₯))/(𝑔(π‘₯)) = ((π‘™π‘–π‘š)┬(π‘₯β†’π‘Ž) 𝑓(π‘₯))/((π‘™π‘–π‘š)┬(π‘₯β†’π‘Ž) 𝑔(π‘₯)) (lim┬(xβ†’1) 𝑓(π‘₯) βˆ’ 2)/(lim┬(xβ†’1) γ€–(π‘₯γ€—^2 βˆ’ 1) ) = Ο€ lim┬(xβ†’1) (f(x) – 2) = Ο€ Γ— lim┬(xβ†’1) (x2 – 1) lim┬(xβ†’1) f(x) – lim┬(xβ†’1) 2 = Ο€ (lim┬(xβ†’1) x2 – lim┬(xβ†’1) 1) Finding limits, putting x = 1 lim┬(xβ†’1) f(x) – 2 = Ο€ Γ— ((1)2 – 1) lim┬(xβ†’1) f(x) – 2 = Ο€ Γ— 0 lim┬(xβ†’1) f(x) – 2 = Ο€ Γ— 0 lim┬(xβ†’1) f(x) – 2 = 0 lim┬(xβ†’1) f(x) = 2 Thus (π’π’Šπ’Ž)┬(π±β†’πŸ) f (x) = 2

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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.