. Evaluate: lim┬(x→π)〖(x/(x-π) ∫sint/dt from x to π ) 〗
(lim as x approaches to π)
Method - 1
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L = lim (x→π) { x. ∫ [x,π] sin t dt } / ( π - x ) ........ 0/0 ... use L'HR
= lim ( x → π) { x. (-sin x) + 1. ∫ [x,π] sin t dt } / (-1)
= { π(-sin π) + 1. ∫ [π,π] sin t dt } / (-1)
= { π(0) + 1(0) } / (-1)
= 0/(-1)
= 0 .................. Ans.
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Method - 2
∫ [x,π] sin t dt = - [ cos t ] = - [ cos π - cos x ] = - ( -1 - cos x ) = 1 + cos x.
Then, the required limit is
= lim (x→π) [ x ( 1 + cos x ) / ( π - x ) ]
= lim (x→π) x • lim (x→π) { [ 1 - cos (π-x) ] / (π-x) }
= π • lim ( θ → 0 ) [ ( 1 - cos θ ) / ( θ ) ], ... where θ = π - x
= π • [ 0 ]
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Method - 1
▬▬▬▬▬▬
L = lim (x→π) { x. ∫ [x,π] sin t dt } / ( π - x ) ........ 0/0 ... use L'HR
= lim ( x → π) { x. (-sin x) + 1. ∫ [x,π] sin t dt } / (-1)
= { π(-sin π) + 1. ∫ [π,π] sin t dt } / (-1)
= { π(0) + 1(0) } / (-1)
= 0/(-1)
= 0 .................. Ans.
..........................................................................................
Method - 2
▬▬▬▬▬▬
∫ [x,π] sin t dt = - [ cos t ] = - [ cos π - cos x ] = - ( -1 - cos x ) = 1 + cos x.
Then, the required limit is
= lim (x→π) [ x ( 1 + cos x ) / ( π - x ) ]
= lim (x→π) x • lim (x→π) { [ 1 - cos (π-x) ] / (π-x) }
= π • lim ( θ → 0 ) [ ( 1 - cos θ ) / ( θ ) ], ... where θ = π - x
= π • [ 0 ]
= 0 .................. Ans.
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Happy To Help !
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