Chapter 13 Class 11 Limits and Derivatives
Example 3 (i) Important
Ex 12.1, 6 Important
Ex 12.1,10 Important You are here
Ex 12.1, 13
Ex 12.1, 16
Ex 12.1, 22 Important
Ex 12.1, 25 Important
Ex 12.1, 28 Important
Ex 12.1, 30 Important
Ex 12.1, 32 Important
Ex 12.2, 9 (i)
Ex 12.2, 11 (i)
Example 20 (i)
Example 21 (i)
Example 22 (i)
Misc 1 (i)
Misc 6 Important
Misc 9 Important
Misc 24 Important
Misc 27 Important
Misc 28 Important
Misc 30 Important
Chapter 13 Class 11 Limits and Derivatives
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