Is there a way to compute $\lim\limits_{x\to\pi/3}\frac{\sqrt{3+2\cos x}-2}{\ln(1+\sin3x)}$ without using L'hopital? I can compute 
$$\lim_{x\to\pi/3}\frac{\sqrt{3+2\cos x}-2}{\ln(1+\sin3x)}$$ using L'hopital and the limit equals $\frac{\sqrt{3}}{12}$, but is there another way to compute this limit without using L'hopital, please do it for me if there is.
Many thanks.

$$\lim_{x\to\pi/3}\frac{\sqrt{3+2\cos x}-2}{\ln(1+\sin3x)}=\lim_{x\to\pi/3}\frac{2\cos x-1}{\ln(1+\sin3x)}\frac{1}{\sqrt{3+2\cos x}+2}$$
$$=\frac{1}{4}\lim_{x\to\pi/3}\frac{2\cos x-1}{\ln(1+\sin3x)}=\frac{1}{4}\lim_{x\to\pi/3}\frac{2\cos x-1}{\sin(3x)}\frac{\sin(3x)}{\ln(1+\sin3x)}$$
$\lim_{x\to\pi/3}\frac{2\cos x-1}{\sin(3x)}$ is just a standard trig limit, while
$$\lim_{x\to\pi/3}\frac{\sin(3x)}{\ln(1+\sin3x)}=\frac{1}{\lim_{x\to\pi/3}\frac{\ln(1+\sin3x)}{\sin(3x)}}=\frac{1}{\lim_{y \to 1}\frac{\ln(y)}{y-1}}=\frac{1}{\ln'(1)}=1$$
