Let x=π3+t
limt→0sin(π3−π3−t)2cos(π3+t)−1
=limt→0−sint2[cos(t+π3]−1
=limt→0−sint2[12cost−√32sint]−1
=limt→0−sintcost−√3sint−1
=limt→0+sint√3sint+1−cost
=limt→02sint2cost2√3×2sint2cost2+2sin2t2
=limt→02sint2(cost2)2sint2(√3cost2+sint2)
=limt→0cost2√3cost2+sint2
⇒1√3+0=1√3.