# Problem in the properties of limit: $\lim\limits_{x \to\frac{\pi}{3}}\frac{\sin\left(x-\frac{\pi}{3}\right)}{1-2\cos\left(x\right)}$

$$\lim\limits_{x \to\frac{\pi}{3}}\frac{\sin\left(x-\frac{\pi}{3}\right)}{1-2\cos\left(x\right)}$$

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?

## 1 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

$$\frac{1-\cos u}{\sin u}=u\frac{1-\cos u}{u^2}\frac{u}{\sin u}\to 0$$

• I think that the identity $\frac{1-\cos u}{\sin u}=\frac{\sin u}{1+\cos u}$ used by OP suits much better to the required approach (no Taylor series or L'Hopital rule). – user Nov 23 '19 at 10:15
• @user My approach also is without l’Hospital and Taylor indeed it makes use only of standard limits. – user Nov 23 '19 at 10:43