Ex 12.1, 8 - Chapter 12 Class 11 Limits and Derivatives

Last updated at Dec. 16, 2024 by

Slide11.JPG

Slide12.JPG
Slide13.JPG

Go Ad-free

Transcript

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

Davneet Singh's photo - Co-founder, Teachoo

Made by

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