$$\lim\limits_{x \to\frac{\pi}{3}}\frac{\sin\left(x-\frac{\pi}{3}\right)}{1-2\cos\left(x\right)}$$
I used the following property: if $$\lim\limits_{\large x \to\frac{\pi}{3}}f(x)=L$$ then $$\lim\limits_{x \to\frac{\pi}{3}}\frac{1}{f\left(x\right)}=\frac{1}{L}$$
where $L$ is a real number and nonzero,hence we have:
$$\lim\limits_{\large x \to\frac{\pi}{3}}\frac{1-2\cos\left(x\right)}{\sin\left(x-\frac{\pi}{3}\right)}$$
substititute $x-\frac{\pi}{3}=u$:
$$\lim\limits_{\large u \to 0}\frac{1-2\cos\left(u+\frac{\pi}{3}\right)}{\sin\left(u\right)}$$$$=\lim\limits_{\large u \to 0}\frac{1-\cos\left(u\right)+\sqrt{2}\sin\left(u\right)}{\sin\left(u\right)}=\lim\limits_{\large u \to 0}\frac{1-\cos\left(u\right)}{\sin\left(u\right)}+\sqrt{2}$$$$=\lim\limits_{\large u \to 0}\frac{\sin\left(u\right)}{1+\cos\left(u\right)}+\sqrt{2}=\sqrt{2}$$
hence the main limit should be $\frac{1}{\sqrt{2}}$which is wrong, but I don't know why, also is there any way to solve the problem without using Taylor series or L'hopital's rule?