$$\lim\limits_{x \to\frac{\pi}{3}}\frac{\sin\left(x-\frac{\pi}{3}\right)}{1-2\cos\left(x\right)}$$
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?
Best Answer
Your derivation is absolutely fine and right, but we have that
$$1-2\cos\left(u+\frac{\pi}{3}\right)=1-2\frac12\cos u+2\frac {\sqrt 3} 2\sin u=1-\cos u+\color{red}{\sqrt 3}\sin u$$
therefore
$$\lim\limits_{x \to\frac{\pi}{3}}\frac{\sin\left(x-\frac{\pi}{3}\right)}{1-2\cos\left(x\right)}=\frac1{\sqrt 3}$$
Note also that we don't need to invert the expression, indeed in the same way we have
$$\lim\limits_{\large u \to 0}\frac{\sin\left(u\right)}{1-2\cos\left(u+\frac{\pi}{3}\right)}=\lim\limits_{\large u \to 0}\frac{\sin\left(u\right)}{1-\cos u+\sqrt 3\sin u}=\lim\limits_{\large u \to 0}\frac{1}{\frac{1-\cos u}{\sin u}+\sqrt 3}=\frac1{\sqrt 3}$$
since
$$\frac{1-\cos u}{\sin u}=u\frac{1-\cos u}{u^2}\frac{u}{\sin u}\to 0$$