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Vinay Arya Grade: 12
        

 Q.1. Find Limitx---->0 8/x8[1-cosx2/2 -cosx2/4 +cosx2/2 .cosx2/4]

8 years ago

Answers : (1)

askIITiansexpert nagesh
16 Points
										

 


Limitx---->08/x8[(1-cosx2/2)(1-cosx2/4)]=Limitx--->08[(1-cosx2/2)/x4][(1-cosx2/4)/x4]


=Limitx---->08[2(sinx2/4)2/x4][2(sinx2/8)2/x4]=Limitx---->032[(sinx2/4)/4(x2/4)]2[(sinx2/8)/8(x2/8)]2


=Limitx--->032/(4282) [(sinx2/4)/(x2/4)]2[(sinx2/8)/(x2/8)]2  =1/32*Limitx---->0[(sinx2/4)/(x2/4)]2[(sinx2/8)/(x2/8)]2 


=1/32


Here i used following formulas:


1-cosx=2sin2(x/2)


Limity--->0siny/y = 1 [replace y by x2/4 and x2/8 to get a new result which is used in solving above problem...]


 


 


 

8 years ago
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