`lim_(x-> ppi/3)(tan^3 x - 3tanx)/ cos(x+ pi/6)= 0/0```

SUBSTITUTION FAILED, Then, we will simplify.

==> We know that cos(a+b)= cosa*cosb - sina*sinb

==.> (cos(x+pi/6)= cosx*cos pi/6 - sinx*sin(pi/6)

==>` cos(x+ pi/6)= (sqrt3 cosx)/2 - sinx /2 = (sqrt3 cosx - sinx)/2 .........(1)```

`==> tan^3 x - 3tanx = tanx(tan^2 x - 3) = tanx(tanx -sqrt3)(tanx+sqrt3)`

`==> tanx(tan^2 x - 3)= tanx (sinx/cosx - sqrt3)(sinx/cosx +sqrt3)`

` = tanx(( sinx -sqrt3 cosx)/cosx)((sinx +sqrt3 cosx)/cosx)`

`==> lim_(x-> pi/3) (2tanx(sinx-sqrt3 cosx)(sinx+sqrt3 cosx))/ ((cos^2 x)(sqrt3 cosx - sinx))`

`==> lim_(x-> pi/3) ((-2tanx)(sinx+sqrt3 cosx))/(cos^2 x)`

`= ((-2sqrt3)(sqrt3/2 + sqrt3/2))/ (1/2)^2 = (-2*3) / (1/4) = -6*4 = -24`

`==> lim_(x->pi/3) (tan^3 x- 3tanx) / (cos(x+pi/6)) = -24`

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