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Ex 13.1, 10 - Evaluate: lim z->1 z^1/3 - 1 / z^1/6 - 1 - Teachoo

Ex 13.1,10 - Chapter 13 Class 11 Limits and Derivatives - Part 2
Ex 13.1,10 - Chapter 13 Class 11 Limits and Derivatives - Part 3
Ex 13.1,10 - Chapter 13 Class 11 Limits and Derivatives - Part 4

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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

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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.