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Question Number 34522 by rahul 19 last updated on 07/May/18
limx→0loge{sin(a+1x)sina}x,0<a<π2.
Commented byrahul 19 last updated on 08/May/18
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Commented bymath khazana by abdo last updated on 09/May/18
letputA(x)=ln{sin(a+1x)sina}xwehave A(x)=x{ln(a+1x)−ln(sina)} =x{ln(1+ax)−lnx−ln(sina)} =xln(1+ax)−xlnx−xln(sina)⇒ limx→0A(x)=limx→0xln(1+ax)but ln(1+ax)∼ax⇒limx→0A(x)=limx→0ax2=0
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