Examples

Chapter 12 Class 11 Limits and Derivatives
Serial order wise

### Transcript

Example 3 Evaluate: (i) (πππ)β¬(π₯β1) (π₯ 15 β 1)/(π₯10 β 1) (πππ)β¬(π₯β1) (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 (πππ)β¬(π₯βπ) (π₯^π β π^π)/(π₯ β π) = na n β 1 Hence, (πππ)β¬(π₯β1) (π₯^15 β 1)/(π₯^10 β 1) = (πππ)β¬(π₯β1) π₯^15 β 1 Γ·limβ¬(xβ1) x10 β 1 = (πππ)β¬(π₯β1) π₯^15 β γ(1)γ^15 Γ· limβ¬(xβ1) x10 β (1)10 Multiplying and dividing by x β 1 = (πππ)β¬(π₯β1) (π₯^15 β 1^15)/(π₯ β 1) Γ· (πππ)β¬(π§β1) (π₯^10 β γ(10)γ^10)/(π₯ β 1) Using (πππ)β¬(π₯βπ) ( π₯^π β π^π)/(π₯ β π) = nan β 1 Using (πππ)β¬(π₯βπ) ( π₯^π β π^π)/(π₯ β π) = nan β 1 (πππ)β¬(π₯β1) (π₯^15 β γ(1)γ^15)/(π₯ β 1) = 15(1)15 β 1 = 15 (1)14 = 15 (πππ)β¬(π₯β1) (π₯^10 β γ(1)γ^10)/(π₯ β 1) = 10(1)10 β 1 = 10 (1)9 = 10 Hence , (πππ)β¬(π₯β1) (π₯^15 β 1^15)/(π₯ β 1) Γ· (πππ)β¬(π₯β1) (π₯^10 β110)/(π₯ β 1) = 15 Γ· 10 = 15/10 = 3/2 β΄ (πππ)β¬(πβπ) (π^ππ β π)/(π^ππ β π) = π/π