Ex 12.1

Chapter 12 Class 11 Limits and Derivatives
Serial order wise

### Transcript

Ex 12.1, 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 it by using (πππ)β¬(π₯βπ) (π₯^π β π^π)/(π₯ β π) = nan 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) Using (πππ)β¬(π₯βπ) ( π₯^π β π^π)/(π₯ β π) = nan β 1 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 limβ¬(zβ1) (π§^(1/6) β γ(1)γ^(1/6))/(π§ β1) = 1/6 γ(1)γ^(1/6 β 1) = 1/6 Γ 1 = 1/6 Hence our equation becomes = 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