`lim_(x->0)[(1-cos2x)(3+cosx)]/[xtan4x]` is equal to:

Explanation

`lim_(x->0)((1-cos2x)(3+cosx))/(xtan4x)`

`=lim_(x->0)((1-(1-2sin^2 x))(3+cosx))/(xtan4x)`

`=lim_(x->0)((2sinx)/x)(sinx)(3+cosx)/(tan4x)`

`=lim_(x->0)((2sinx)/x)(sinx)(3+cosx)/(4x(tan4x)/(4x)`

We know, `lim_(x->0) sinx/x = 1` and `lim_(x->0) tanx/x = 1`

`:.` our expression becomes,

`=2/4 lim_(x->0)(3+4cosx)`

As `cos 0 = 1`

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