Ex 12.1
Ex 12.1, 2
Ex 12.1, 3
Ex 12.1, 4 Important
Ex 12.1, 5
Ex 12.1, 6 Important
Ex 12.1, 7
Ex 12.1, 8 Important
Ex 12.1, 9
Ex 12.1,10 Important You are here
Ex 12.1, 11
Ex 12.1, 12
Ex 12.1, 13
Ex 12.1, 14 Important
Ex 12.1, 15 Important
Ex 12.1, 16
Ex 12.1, 17 Important
Ex 12.1, 18
Ex 12.1, 19 Important
Ex 12.1, 20
Ex 12.1, 21 Important
Ex 12.1, 22 Important
Ex 12.1, 23
Ex 12.1, 24
Ex 12.1, 25 Important
Ex 12.1, 26
Ex 12.1, 27
Ex 12.1, 28 Important
Ex 12.1, 29
Ex 12.1, 30 Important
Ex 12.1, 31
Ex 12.1, 32 Important
Last updated at May 7, 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