Slide14.JPG

Slide15.JPG
Slide16.JPG Slide17.JPG

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

Transcript

Ex 13.1, 10 Evaluate the Given limit: lim┬(zβ†’1) (𝑧^(1/3) βˆ’ 1)/(𝑧^(1/6) βˆ’ 1) lim┬(zβ†’1) (𝑧^(1/3) βˆ’ 1)/(𝑧^(1/6) βˆ’ 1) = (γ€–(1)γ€—^(1/3) βˆ’ 1)/(γ€–(1)γ€—^(1/6) βˆ’ 1) = (1 βˆ’ 1)/(1 βˆ’ 1) = 0/0 Since it is form 0/0, We can solve it by using (π‘™π‘–π‘š)┬(π‘₯β†’π‘Ž) (π‘₯^𝑛 βˆ’ π‘Ž^𝑛)/(π‘₯ βˆ’ π‘Ž) = nan – 1 Hence, lim┬(zβ†’1) (𝑧^(1/3) βˆ’ 1)/(𝑧^(1/6) βˆ’ 1) = lim┬(zβ†’1) 𝑧^(1/3) – 1 Γ· lim┬(zβ†’1) 𝑧^(1/6) βˆ’ 1 = lim┬(zβ†’1) 𝑧^(1/3) – γ€–(1)γ€—^(1/3) Γ· lim┬(zβ†’1) 𝑧^(1/6) – γ€–(1)γ€—^(1/6) Multiplying and dividing by z – 1 = lim┬(zβ†’1) (𝑧^(1/3) βˆ’ γ€–(1)γ€—^(1/3))/(𝑧 βˆ’ 1) Γ· lim┬(zβ†’1) (𝑧^(1/6) βˆ’γ€– (1)γ€—^(1/6))/(𝑧 βˆ’ 1) Using (π‘™π‘–π‘š)┬(π‘₯β†’π‘Ž) ( π‘₯^𝑛 βˆ’ π‘Ž^𝑛)/(π‘₯ βˆ’ π‘Ž) = nan – 1 lim┬(zβ†’1) (𝑧^(1/3) βˆ’ γ€–(1)γ€—^(1/3))/(𝑧 βˆ’1) = 1/3 γ€–(1)γ€—^(1/3 βˆ’ 1) = 1/3 Γ— 1 = 1/3 lim┬(zβ†’1) (𝑧^(1/6) βˆ’ γ€–(1)γ€—^(1/6))/(𝑧 βˆ’1) = 1/6 γ€–(1)γ€—^(1/6 βˆ’ 1) = 1/6 Γ— 1 = 1/6 Hence our equation becomes = lim┬(zβ†’1) (𝑧^(1/3) βˆ’ γ€–(1)γ€—^(1/3))/(𝑧 βˆ’ 1) Γ· lim┬(zβ†’1) (𝑧^(1/6) βˆ’ 6)/(𝑧 βˆ’ 1) = 1/3 Γ·1/6 = 1/3 Γ— 6/1 = 2

About the Author

Davneet Singh's photo - Teacher, Computer Engineer, Marketer
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.