Last updated at Dec. 16, 2024 by Teachoo
Ex 12.1, 10 Evaluate the Given limit: limβ¬(zβ1) (π§^(1/3) β 1)/(π§^(1/6) β 1) limβ¬(zβ1) (π§^(1/3) β 1)/(π§^(1/6) β 1) = (γ(1)γ^(1/3) β 1)/(γ(1)γ^(1/6) β 1) = (1 β 1)/(1 β 1) = 0/0 Since it is form 0/0, We can solve it by using (πππ)β¬(π₯βπ) (π₯^π β π^π)/(π₯ β π) = nan β 1 Hence, limβ¬(zβ1) (π§^(1/3) β 1)/(π§^(1/6) β 1) = limβ¬(zβ1) π§^(1/3) β 1 Γ· limβ¬(zβ1) π§^(1/6) β 1 = limβ¬(zβ1) π§^(1/3) β γ(1)γ^(1/3) Γ· limβ¬(zβ1) π§^(1/6) β γ(1)γ^(1/6) Multiplying and dividing by z β 1 = limβ¬(zβ1) (π§^(1/3) β γ(1)γ^(1/3))/(π§ β 1) Γ· limβ¬(zβ1) (π§^(1/6) βγ (1)γ^(1/6))/(π§ β 1) Using (πππ)β¬(π₯βπ) ( π₯^π β π^π)/(π₯ β π) = nan β 1 limβ¬(zβ1) (π§^(1/3) β γ(1)γ^(1/3))/(π§ β1) = 1/3 γ(1)γ^(1/3 β 1) = 1/3 Γ 1 = 1/3 limβ¬(zβ1) (π§^(1/6) β γ(1)γ^(1/6))/(π§ β1) = 1/6 γ(1)γ^(1/6 β 1) = 1/6 Γ 1 = 1/6 Hence our equation becomes = limβ¬(zβ1) (π§^(1/3) β γ(1)γ^(1/3))/(π§ β 1) Γ· limβ¬(zβ1) (π§^(1/6) β 6)/(π§ β 1) = 1/3 Γ·1/6 = 1/3 Γ 6/1 = 2
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About the Author
Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo