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

Example  2 - Chapter 13 Class 11 Limits and Derivatives - Part 9
Example  2 - Chapter 13 Class 11 Limits and Derivatives - Part 10


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 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.