Examples

Chapter 12 Class 11 Limits and Derivatives
Serial order wise

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class

### 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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.