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

Transcript

Example 2 Find the limits: (i) (π‘™π‘–π‘š)┬(π‘₯β†’1) [(π‘₯2 + 1)/(π‘₯ + 100)] (π‘™π‘–π‘š)┬(π‘₯β†’1) [(π‘₯2 + 1)/(π‘₯ + 100)] Putting x = 1 = (12 + 1)/(1 + 100) = (1 + 1 )/101 = 𝟐/𝟏𝟎𝟏 Example 2 Find the limits: (ii) (π‘™π‘–π‘š)┬(π‘₯β†’2) [(π‘₯^3 βˆ’ 4π‘₯^2 + 4π‘₯)/(π‘₯^2 βˆ’ 4)] (π‘™π‘–π‘š)┬(π‘₯β†’2) [(π‘₯^3 βˆ’ 4π‘₯^2 + 4π‘₯)/(π‘₯^2 βˆ’ 4)] = lim┬(xβ†’2) π‘₯(π‘₯^2 βˆ’ 4π‘₯ + 4)/(π‘₯^2 βˆ’γ€– (2)γ€—^2 ) = lim┬(xβ†’2) (π‘₯ (π‘₯^2 + 2^2 βˆ’ 4π‘₯ ))/((π‘₯ βˆ’ 2) (π‘₯ + 2)) = lim┬(xβ†’2) (π‘₯ (π‘₯ βˆ’ 2)^2)/((π‘₯ βˆ’ 2) (π‘₯ + 2)) = (π’π’Šπ’Ž)┬(π’™β†’πŸ) 𝒙(𝒙 βˆ’ 𝟐)/(𝒙 + 𝟐) Putting x = 2 = (2 (2 βˆ’ 2))/(2 + 2) = (2 (0))/4 = 0/4 = 0 Example 2 Find the limits: (iii) lim┬(xβ†’2) [(x2 βˆ’4)/(x3 βˆ’ 4x2 + 4x)] lim┬(xβ†’2) [(x2 βˆ’ 4)/(x3βˆ’ 4x2 + 4x)] = lim┬(xβ†’2) [(x2 βˆ’(2)2)/(x(x2 βˆ’ 4x + 4) )] = lim┬(xβ†’2) [((π‘₯ βˆ’ 2) (π‘₯ + 2))/(x(x(x βˆ’ 2) βˆ’ 2(x βˆ’ 2)) )] = lim┬(xβ†’2) [((π‘₯ βˆ’ 2) (π‘₯ + 2))/(π‘₯ (x βˆ’ 2)(x βˆ’ 2) )] = (π₯𝐒𝐦)┬(π±β†’πŸ) [(𝒙 + 𝟐)/(𝐱 (𝐱 βˆ’ 𝟐) )] Putting x = 2 = (2 + 2)/(2 (2 βˆ’ 2) ) = (2 + 2)/(2 (2 βˆ’ 2) ) = 4/(2(0)) = πŸ’/𝟎 = ∞ Which is not defined Example 2 Find the limits: (iv) lim┬(xβ†’2) [(x3 βˆ’2π‘₯2)/(x2βˆ’5x+6)] lim┬(xβ†’2) [(x3 βˆ’ 2π‘₯2)/(x2 βˆ’ 5x + 6)] = lim┬(xβ†’2) ((x2 (x βˆ’ 2))/(π‘₯2 βˆ’ 3π‘₯ βˆ’ 2π‘₯ + 6)) = lim┬(xβ†’2) ((x2 (x βˆ’ 2))/(x (x βˆ’ 3) βˆ’ 2 (x βˆ’ 3) )) = lim┬(xβ†’2) ((x2 (x βˆ’ 2))/((x βˆ’ 2) (x βˆ’ 3) )) = (π₯𝐒𝐦)┬(π±β†’πŸ) ((𝐱𝟐 )/((𝐱 βˆ’ πŸ‘) )) Putting x = 2 = (2)2/(2 βˆ’ 3) = (2)2/(2 βˆ’ 3) = 4/(βˆ’1) = – 4 Example 2 Find the limits: (v) lim┬(xβ†’1) [(x βˆ’2)/(x2βˆ’x)βˆ’ 1/(π‘₯3 βˆ’3π‘₯2+2π‘₯)] lim┬(xβ†’1) [(x βˆ’ 2)/(x2 βˆ’ x)βˆ’ 1/(π‘₯3 βˆ’ 3π‘₯2 + 2π‘₯)] = lim┬(xβ†’1) [(x βˆ’ 2)/(x (x βˆ’1))βˆ’ 1/(π‘₯ (π‘₯2 βˆ’ 3π‘₯ + 2))] = lim┬(xβ†’1) [(x βˆ’ 2)/(x (x βˆ’1))βˆ’ 1/(π‘₯ (π‘₯2 βˆ’ 2π‘₯ βˆ’ π‘₯ + 2))] = lim┬(xβ†’1) [(x βˆ’ 2)/(x (x βˆ’1))βˆ’ 1/(π‘₯ (π‘₯ (π‘₯ βˆ’ 2) βˆ’ 1 (π‘₯ βˆ’ 2)))] = lim┬(xβ†’1) [(x βˆ’ 2)/(x (x βˆ’1))βˆ’ 1/(π‘₯ (π‘₯ βˆ’ 1) (π‘₯ βˆ’ 2))] = lim┬(xβ†’1) [((x βˆ’ 2)(π‘₯ βˆ’ 2) βˆ’ 1)/(π‘₯ (π‘₯ βˆ’ 1) (π‘₯ βˆ’ 2))] = lim┬(xβ†’1) [((x βˆ’ 2)2 βˆ’ 1)/(π‘₯ (π‘₯ βˆ’ 1) (π‘₯ βˆ’ 2))] = lim┬(xβ†’1) [(π‘₯2 + 2^2 βˆ’ 4π‘₯ βˆ’ 1)/(π‘₯ (π‘₯ βˆ’ 1) (π‘₯ βˆ’ 2))] = lim┬(xβ†’1) [(π‘₯2 + 4 βˆ’ 4π‘₯ βˆ’ 1)/(π‘₯ (π‘₯ βˆ’ 1) (π‘₯ βˆ’ 2))] = (π₯𝐒𝐦)┬(π’™β†’πŸ) [(π’™πŸ βˆ’ πŸ’π’™ βˆ’ πŸ‘)/(𝒙 (𝒙 βˆ’ 𝟏) (𝒙 βˆ’ 𝟐))] Putting x = 1 = (1^2 βˆ’ 4 (1) + 3)/(1(1 βˆ’ 1)(1 βˆ’ 2)) = 𝟎/𝟎 Since it is 0/0 form we can simplify = lim┬(xβ†’1) [(π‘₯2 βˆ’ 4π‘₯ βˆ’ 3)/(π‘₯ (π‘₯ βˆ’ 1) (π‘₯ βˆ’ 2))] = lim┬(xβ†’1) [(π‘₯2 βˆ’ 3π‘₯ βˆ’ π‘₯ + 3)/(π‘₯ (π‘₯ βˆ’ 1) (π‘₯ βˆ’ 2))] = lim┬(xβ†’1) [(π‘₯ (π‘₯ βˆ’ 3) βˆ’ 1(π‘₯ βˆ’ 3))/(π‘₯ (π‘₯ βˆ’ 1) (π‘₯ βˆ’ 2))] = lim┬(xβ†’1) [((π‘₯ βˆ’ 1) (π‘₯ βˆ’ 3))/(π‘₯ (π‘₯ βˆ’ 1) (π‘₯ βˆ’ 2))] = (π₯𝐒𝐦)┬(π’™β†’πŸ) [( 𝒙 βˆ’ πŸ‘)/(𝒙 (𝒙 βˆ’ 𝟐))] Putting x = 1 = (1 βˆ’ 3)/(1 (1 βˆ’ 2)) = (βˆ’2)/(1 Γ— βˆ’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 9 years. He provides courses for Maths and Science at Teachoo.