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Last updated at Sept. 6, 2021 by Teachoo

Maths Crash Course - Live lectures + all videos + Real time Doubt solving!

Example 3 Evaluate: (ii) (πππ)β¬(π₯β0) (β(1 + π₯) β 1)/π₯ (πππ)β¬(π₯β0) (β(1 + x )β 1)/x Putting x = 0 = (β(1 + 0) β 1)/0 = (β(1 ) β 1)/0 = (1 β 1)/0 = 0/0 Since it is a 0/0 form We simplify the equation Putting y = 1 + x β y β 1 = x As x β 0 y β 1 + 0 y β 1 So, our equation becomes (πππ)β¬(π₯β0) (β(1 + π₯ )β 1)/π₯ = (πππ)β¬(π¦β1) (βπ¦ β 1)/(π¦ β 1) = (πππ)β¬(π₯β1) ( π¦^((β1)/2) β 1)/(π¦ β 1) = (πππ)β¬(π₯β1) ( π¦^((β1)/2) β 1^((β1)/2))/(π¦ β 1) = 1/2 Γ 1^((β1)/2 β 1) = 1/2 Γ 1 = π/π