Chapter 13 Class 11 Limits and Derivatives
Example 3 (i) Important You are here
Ex 12.1, 6 Important
Ex 12.1,10 Important
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 Dec. 16, 2024 by Teachoo
Example 3 Evaluate: (i) (πππ)β¬(π₯β1) (π₯ 15 β 1)/(π₯10 β 1) (πππ)β¬(π₯β1) (π₯ 15 β 1)/(π₯10 β 1) = (γ(1)γ^15 β 1)/(γ(1)γ^10 β 1) = (1 β 1)/(1 β 1) = 0/0 Since it is form 0/0, We can solve by using theorem (πππ)β¬(π₯βπ) (π₯^π β π^π)/(π₯ β π) = na n β 1 Hence, (πππ)β¬(π₯β1) (π₯^15 β 1)/(π₯^10 β 1) = (πππ)β¬(π₯β1) π₯^15 β 1 Γ·limβ¬(xβ1) x10 β 1 = (πππ)β¬(π₯β1) π₯^15 β γ(1)γ^15 Γ· limβ¬(xβ1) x10 β (1)10 Multiplying and dividing by x β 1 = (πππ)β¬(π₯β1) (π₯^15 β 1^15)/(π₯ β 1) Γ· (πππ)β¬(π§β1) (π₯^10 β γ(10)γ^10)/(π₯ β 1) Using (πππ)β¬(π₯βπ) ( π₯^π β π^π)/(π₯ β π) = nan β 1 Using (πππ)β¬(π₯βπ) ( π₯^π β π^π)/(π₯ β π) = nan β 1 (πππ)β¬(π₯β1) (π₯^15 β γ(1)γ^15)/(π₯ β 1) = 15(1)15 β 1 = 15 (1)14 = 15 (πππ)β¬(π₯β1) (π₯^10 β γ(1)γ^10)/(π₯ β 1) = 10(1)10 β 1 = 10 (1)9 = 10 Hence , (πππ)β¬(π₯β1) (π₯^15 β 1^15)/(π₯ β 1) Γ· (πππ)β¬(π₯β1) (π₯^10 β110)/(π₯ β 1) = 15 Γ· 10 = 15/10 = 3/2 β΄ (πππ)β¬(πβπ) (π^ππ β π)/(π^ππ β π) = π/π