$\underset{x→\frac{π}{3}}{\lim}\frac{\sin(\frac{π}{3}-x)}{2\cos x-1}$ is equal to:

$\frac{1}{\sqrt{3}}$

$L=\underset{x→\frac{π}{3}}{\lim}\frac{\sin(\frac{π}{3}-x)}{2\cos x-1}$

Put $\frac{π}{3}-x=y$

$L=\underset{y→0}{\lim}\frac{\sin y}{\cos(\frac{π}{3}-y)-1}=\underset{y→0}{\lim}\frac{\sin y}{2×\frac{1}{2}\cos y+2×\frac{\sqrt{3}}{2}\sin y -1}=\underset{y→0}{\lim}\frac{\sin y}{\cos y+\sqrt{3}\sin y-1}$

$\underset{y→0}{\lim}\frac{\sin y}{\sqrt{3}\sin y-2\sin^2\frac{y}{2}}(1-\cos y=2\sin^2\frac{y}{2})=\underset{y→0}{\lim}\frac{\cos \frac{y}{2}}{\sqrt{3}\cos \frac{y}{2}-\sin (\frac{y}{2})}(\sin y=2\sin\frac{y}{2}\cos\frac{y}{2})=\frac{1}{\sqrt{3}}$