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

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

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

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

=1/32

Here i used following formulas:

1-cosx=2sin²(x/2)

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