Put, x=2π−h in the given limit, we get
=h→0lim[(π−2×(2π−h))3cot(2π−h)−cos(2π−h)]
=h→0lim(π−2×2π+2h)3tanh−sinh
=h→0lim[8h3tanh−sinh]
=81h→0limhsinh[h2sech−1]
=81[h→0lim(hsinh)][h→0lim((cosh)h21−cosh)]
=81×1×[h→0lim((cosh)h21−cosh)]
=81h→0lim[(cosh)h22sin2(2h)]
=81h→0lim[(4h2)×4×cosh2sin2(2h)]
=161.