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Let u = a − x . Then 2 a π x = 2 a π ( a − u ) = 2 π − 2 a π u and u → 0 as x → a .
The given limit can then be written as
u → 0 lim u tan ( 2 π − 2 a π u ) = u → 0 lim u cot ( 2 a π u ) = u → 0 lim tan ( 2 a π u ) π 2 a × 2 a π u = π 2 a y → 0 lim tan ( y ) y ,
where y = 2 a π u → 0 as u → 0 . Now y → 0 lim y tan ( y ) = 1 , so the desired limit is just π 2 a .