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Slide67.JPG

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

Transcript

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
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.