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Ex 13.1, 8 - Chapter 13 Class 11 Limits and Derivatives (Term 1 and Term 2)

Last updated at Nov. 30, 2019 by

Ex 13.1, 8 - Evaluate limit lim x->3 (x4 - 81/2x2-5x-3) - Ex 13.1

Ex 13.1, 8 - Chapter 13 Class 11 Limits and Derivatives - Part 2
Ex 13.1, 8 - Chapter 13 Class 11 Limits and Derivatives - Part 3

Transcript

Ex 13.1, 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

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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.