For the function to be continuous at x=0:
limx→016x21−cos8x=k
Using the trigonometric identity 1−cosθ=2sin2(2θ) with θ=8x:
1−cos8x=2sin2(4x)
limx→016x21−cos8x
=limx→016x22sin2(4x)
=limx→08x2sin2(4x)
Rewriting the expression:
8x2sin2(4x)
=(4x)2sin2(4x)×8x2(4x)2
=[4xsin(4x)]2×8x216x2
=[4xsin(4x)]2×2
Using the standard limit limx→0xsinx=1:
limx→04xsin(4x)=1
limx→0[4xsin(4x)]2×2
=(1)2×2
=2
Therefore, k=2