Ex 13.1, 8
Last updated at Dec. 8, 2016 by Teachoo
Transcript
Ex 13.1, 8 Evaluate the Given limit: limx→3 x4 −812x2 −5x−3 limx→3 x4 − 812x2 − 5x − 3 = 34 − 812 32 − 5 3 − 3 = 81 − 8118 − 15 − 3 = 00 Since it is a 00 form we simplify as limx→3 x4 − 812x2 − 5x − 3 = limx→3 (𝑥2)2− (9)22x2 −6x + x − 3 limx→3 x4 − 812x2 − 5x − 3 = limx→3 (𝑥2)2− (9)22x2 −6x + x − 3 = limx→3 x2 − 9 (x2 + 9)2x x − 3 + 1 (x − 3) = limx→3 𝑥2− 32(𝑥2 + 9) x + 1(𝑥 − 3) = limx→3 𝑥 − 3 ( 𝑥 + 3)(𝑥2 + 9) 2𝑥 + 1 (𝑥 − 3) = 𝐥𝐢𝐦𝐱→𝟑 𝒙 + 𝟑 (𝒙𝟐 + 𝟗)𝟐𝒙 + 𝟏 = 3 + 3( 32 + 9)2 ×3 +1 = 6 (9 + 9)6 + 1 = 6(18)7 = 𝟏𝟎𝟖𝟕
Ex 13.1
Ex 13.1, 1
Ex 13.1, 2
Ex 13.1, 3
Ex 13.1, 4
Ex 13.1, 5
Ex 13.1, 6 Important
Ex 13.1, 7
Ex 13.1, 8 You are here
Ex 13.1, 9
Ex 13.1,10 Important
Ex 13.1, 11
Ex 13.1, 12
Ex 13.1, 13 Important
Ex 13.1, 14
Ex 13.1, 15
Ex 13.1, 16 Important
Ex 13.1, 17
Ex 13.1, 18
Ex 13.1, 19
Ex 13.1, 20
Ex 13.1, 21
Ex 13.1, 22 Important
Ex 13.1, 23
Ex 13.1, 24
Ex 13.1, 25 Important
Ex 13.1, 26
Ex 13.1, 27
Ex 13.1, 28 Important
Ex 13.1, 29
Ex 13.1, 30 Important
Ex 13.1, 31
Ex 13.1, 32 Important
About the Author
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 7 years. He provides courses for Mathematics and Science from Class 6 to 12. You can learn personally from here https://www.teachoo.com/premium/maths-and-science-classes/.