Question
[
begin{array}{l}text { & }(1-cos 2 x)=2 sin ^{2} x text { i. } lim _{x rightarrow 0} frac{2 sin ^{2} x cdot(3+cos x)}{x cdot tan 4 x} cdot frac{2 cdot lim _{x rightarrow 0} frac{x^{2} x}{tan 4 x} cdot(3+cos x)}{x} & =frac{2(3+1)}{4}=2end{array}
]

16 (1-cos2x)(3 -cosx) x70 xtan 4x is equal to :
Solution
