Ex 13.1, 8 - Chapter 13 Class 11 Limits and Derivatives
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
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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.