1. Chapter 12 Class 11 Limits and Derivatives
  2. Serial order wise
  3. Ex 12.1
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Ex 12.1, 8 - Chapter 12 Class 11 Limits and Derivatives

Last updated at Dec. 16, 2024 by Teachoo

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Ex 12.1, 8 Evaluate the Given limit: lim┬(x→3) (x4 −81)/(2x2 −5x−3) lim┬(x→3) (x4 − 81)/(2x2 − 5x − 3) Putting x = 3 = ((3)4 − 81)/(2 (3)2 − 5 (3) − 3) = (81 − 81)/(18 − 15 − 3) = 0/0 Since it is a 0/0 form we simplify as lim┬(x→3) (x4 − 81)/(2x2 − 5x − 3) = lim┬(x→3) (〖(𝑥2)〗^2 − 〖(9)〗^2)/(2x2 − 6x + x − 3) = lim┬(x→3) ((x2 − 9) (x2 + 9))/(2x (x − 3) + 1 (x − 3)) = lim┬(x→3) ((𝑥2− (3)2) (𝑥2 + 9))/((x + 1)(𝑥 − 3)) = lim┬(x→3) ((𝑥 − 3) (𝑥 + 3)(𝑥2 + 9))/((2𝑥 + 1) (𝑥 − 3)) = (𝐥𝐢𝐦)┬(𝐱→𝟑) ((𝒙 + 𝟑) (𝒙𝟐 + 𝟗))/(𝟐𝒙 + 𝟏) (Using a2 – b2 = (a – b) (a + b)) (Using a2 – b2 = (a – b) (a + b)) 𝑃𝑢𝑡𝑡𝑖𝑛𝑔 𝑥=3 = ((3 + 3)((3)2 + 9))/(2 ×3 +1) = (6 (9 + 9))/(6 + 1) = (6(18))/7 = 𝟏𝟎𝟖/𝟕
  1. Chapter 12 Class 11 Limits and Derivatives
  2. Serial order wise

    3. Ex 12.1

    Ex 12.2 →
Ex 12.2 →
Chapter 12 Class 11 Limits and Derivatives
Serial order wise

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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