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