Example  2 - Chapter 13 Class 11 Limits and Derivatives - Part 8

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Example  2 - Chapter 13 Class 11 Limits and Derivatives - Part 9

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Example  2 - Chapter 13 Class 11 Limits and Derivatives - Part 10

  1. Chapter 13 Class 11 Limits and Derivatives (Term 1 and Term 2)
  2. Serial order wise

Transcript

Example 2 Find the limits: (v) lim┬(x→1) [(x −2)/(x2−x)− 1/(𝑥3 −3𝑥2+2𝑥)] lim┬(x→1) [(x − 2)/(x2 − x)− 1/(𝑥3 − 3𝑥2 + 2𝑥)] = lim┬(x→1) [(x − 2)/(x (x −1))− 1/(𝑥 (𝑥2 − 3𝑥 + 2))] = lim┬(x→1) [(x − 2)/(x (x −1))− 1/(𝑥 (𝑥2 − 2𝑥 − 𝑥 + 2))] = lim┬(x→1) [(x − 2)/(x (x −1))− 1/(𝑥 (𝑥 (𝑥 − 2) − 1 (𝑥 − 2)))] = lim┬(x→1) [(x − 2)/(x (x −1))− 1/(𝑥 (𝑥 − 1) (𝑥 − 2))] = lim┬(x→1) [((x − 2)(𝑥 − 2) − 1)/(𝑥 (𝑥 − 1) (𝑥 − 2))] = lim┬(x→1) [((x − 2)2 − 1)/(𝑥 (𝑥 − 1) (𝑥 − 2))] = lim┬(x→1) [(𝑥2 + 2^2 − 4𝑥 − 1)/(𝑥 (𝑥 − 1) (𝑥 − 2))] = lim┬(x→1) [(𝑥2 + 4 − 4𝑥 − 1)/(𝑥 (𝑥 − 1) (𝑥 − 2))] = (𝐥𝐢𝐦)┬(𝒙→𝟏) [(𝒙𝟐 − 𝟒𝒙 − 𝟑)/(𝒙 (𝒙 − 𝟏) (𝒙 − 𝟐))] Putting x = 1 = (1^2 − 4 (1) + 3)/(1(1 − 1)(1 − 2)) = 𝟎/𝟎 Since it is 0/0 form we can simplify = lim┬(x→1) [(𝑥2 − 4𝑥 − 3)/(𝑥 (𝑥 − 1) (𝑥 − 2))] = lim┬(x→1) [(𝑥2 − 3𝑥 − 𝑥 + 3)/(𝑥 (𝑥 − 1) (𝑥 − 2))] = lim┬(x→1) [(𝑥 (𝑥 − 3) − 1(𝑥 − 3))/(𝑥 (𝑥 − 1) (𝑥 − 2))] = lim┬(x→1) [((𝑥 − 1) (𝑥 − 3))/(𝑥 (𝑥 − 1) (𝑥 − 2))] = (𝐥𝐢𝐦)┬(𝒙→𝟏) [( 𝒙 − 𝟑)/(𝒙 (𝒙 − 𝟐))] Putting x = 1 = (1 − 3)/(1 (1 − 2)) = (−2)/(1 × −1) = 2

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.