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.

We will have to use some limiting fact about $\log$. We will use that $\displaystyle\lim_{u\to0}\frac{\log(1+u)}{u}=1$.
$$
\begin{align}
\lim_{x\to\pi/3}\frac{\sqrt{3+2\cos(x)}-2}{\log(1+\sin(3x))}
&=\lim_{x\to\pi/3}\frac{\color{#00A000}{\sqrt{3+2\cos(x)}-2}}{\log(1+\sin(3x))}
\frac{\color{#00A000}{\sqrt{3+2\cos(x)}+2}}{\sqrt{3+2\cos(x)}+2}\\
&=\lim_{x\to\pi/3}\frac{\color{#00A000}{2\cos(x)-1}}{\log(1+\sin(3x))}
\color{#C00000}{\frac1{\sqrt{3+2\cos(x)}+2}}\\
&=\color{#C00000}{\frac14}\lim_{x\to\pi/3}\frac{\color{#0000FF}{2\cos(x)-1}}{\log(1+\sin(3x))}
\frac{\color{#0000FF}{(2\cos(x)+1)\sin(x)}}{(2\cos(x)+1)\sin(x)}\\
&=\frac14\lim_{x\to\pi/3}\frac{\color{#0000FF}{\sin(3x)}}{\log(1+\sin(3x))}
\frac1{\color{#00A000}{(2\cos(x)+1)\sin(x)}}\\
&=\frac1{4\color{#00A000}{\sqrt3}}\lim_{x\to\pi/3}\frac{\sin(3x)}{\log(1+\sin(3x))}\\
&=\frac1{4\sqrt3}\lim_{u\to0}\frac{u}{\log(1+u)}\\
&=\frac1{4\sqrt3}
\end{align}
$$

Proof of limit used above
Using $\log(x)=\lim\limits_{n\to\infty}n(x^{1/n}-1)$, which simply inverts $e^x=\lim\limits_{n\to\infty}\left(1+\frac{x}{n}\right)^n$, we have
$$
\begin{align}
\frac{\log(1+u)}{u}
&=\lim_{n\to\infty}n\frac{(1+u)^{1/n}-1}{(1+u)-1}\\
&=\lim_{n\to\infty}\frac{n}{(1+u)^{(n-1)/n}+(1+u)^{(n-2)/n}+\dots+1}
\end{align}
$$
which is easily between $\frac1{1+u}$ and $1$. Thus, the Squeeze Theorem yields
$$
\lim_{u\to0}\frac{\log(1+u)}{u}=1
$$
