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Grade: 12

                        

Q.1. Find Limit x---->0 8/x 8 [1-cosx 2 /2 -cosx 2 /4 +cosx 2 /2 .cosx 2 /4]

11 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...]

 

 

 

11 years ago
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