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Limit x tends to π/2 (Sinx - (sinx)^sinx )/(1-sinx + log sinx)Do it without L hospital and taylorexpansionDo it with basics and properties...

Limit x tends to π/2 (Sinx - (sinx)^sinx )/(1-sinx + log sinx)Do it without L hospital and taylorexpansionDo it with basics and properties...

Grade:11

1 Answers

Nandana
110 Points
4 years ago
hi...
  for this question procede in this way -
Lim x → π/2  (sin x – sinxsin x ) / (1- sin x + log (sin x)) = Lim x → π/2 (1- sinxsin x -1 )/ (1/ sin x  – 1 + log (sin x)/sinx
                                                                                  = 1 – sin xLim x → π/2 ( sin x) / (1/ sin x  – 1 + log (sin x)/sinx
                                                                                  = 1-1 = 0
                                                                 
 

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