1. Class 11
2. Important Question for exams Class 11

Transcript

Ex 13.1, 10 Evaluate the Given limit: limβ¬(zβ1) (π§^(1/3) β 1)/(π§^(1/6) β 1) limβ¬(zβ1) (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 by using theorem limβ¬(xβa) (π₯^π β π^π)/(π₯ β π) = na n β 1 Hence, limβ¬(zβ1) (π§^(1/3) β 1)/(π§^(1/6) β 1) = limβ¬(zβ1) π§^(1/3) β 1 Γ· limβ¬(zβ1) π§^(1/6) β 1 = limβ¬(zβ1) π§^(1/3) β γ(1)γ^(1/3) Γ· limβ¬(zβ1) π§^(1/6) β γ(1)γ^(1/6) Multiplying and dividing by z β 1 = limβ¬(zβ1) (π§^(1/3) β γ(1)γ^(1/3))/(π§ β 1) Γ· limβ¬(zβ1) (π§^(1/6) βγ (1)γ^(1/6))/(π§ β 1) It is of form limβ¬(x β a) (π₯^π β π^π)/(π₯ β π ) = nan β 1 Replacing x with z, a with 1, n with 1/3 limβ¬(zβ1) (π§^(1/3) β γ(1)γ^(1/3))/(π§ β1) = 1/3 γ(1)γ^(1/3 β 1) = 1/3 Γ 1 = 1/3 Similarly replacing x with z, a with 1 and n = 1/6 limβ¬(zβ1) (π§^(1/6) β 1^(1/6))/(π§ β 1) = 1/6 γ(1)γ^(1/6 β 1) = 1/6 Γ 1 = 1/6 Hence our equation = limβ¬(zβ1) (π§^(1/3) β γ(1)γ^(1/3))/(π§ β 1) Γ· limβ¬(zβ1) (π§^(1/6) β 6)/(π§ β 1) = 1/3 Γ·1/6 = 1/3 Γ 6/1 = 2

Class 11
Important Question for exams Class 11