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.
 A: 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
$$
\lim_{u\to0}\frac{\log(1+u)}{u}=1
$$
A: $$\lim_{x\to\pi/3}\frac{\sqrt{3+2\cos x}-2}{\ln(1+\sin3x)}=\lim_{x\to\pi/3}\frac{2\cos x-1}{\ln(1+\sin3x)}\frac{1}{\sqrt{3+2\cos x}+2}$$
$$=\frac{1}{4}\lim_{x\to\pi/3}\frac{2\cos x-1}{\ln(1+\sin3x)}=\frac{1}{4}\lim_{x\to\pi/3}\frac{2\cos x-1}{\sin(3x)}\frac{\sin(3x)}{\ln(1+\sin3x)}$$
$\lim_{x\to\pi/3}\frac{2\cos x-1}{\sin(3x)}$ is just a standard trig limit, while
$$\lim_{x\to\pi/3}\frac{\sin(3x)}{\ln(1+\sin3x)}=\frac{1}{\lim_{x\to\pi/3}\frac{\ln(1+\sin3x)}{\sin(3x)}}=\frac{1}{\lim_{y \to 1}\frac{\ln(y)}{y-1}}=\frac{1}{\ln'(1)}=1$$
A: Let's replace $x$ by $x+\pi/3$ and let $x \to 0$.
I will be as simple-minded as I can.
We use $\sin(\pi/3) = \sqrt{3}/2$ and $\cos(\pi/3) = 1/2$.
$\cos(x+\pi/3) = \cos(x) \cos(\pi/3) - \sin(x)\sin(\pi/3)
= \cos(x)/2 - \sin(x)\sqrt{3}/2
$.
As $x \to 0$,
$\cos(x+\pi/3) \to 1/2$.
$\sin(3(x+\pi/3)) 
= \sin(3x+3\pi/3)
= \sin(3x+\pi)
= \sin(3x)\cos(\pi) + \cos(3x)\sin(\pi)
= -\sin(3x)
\to 0
$.
So, initially, this goes to $0/0$.
Let's use $\sin(x) \approx x$ and
$\cos(x) \approx 1-x^2/2$ as $x \to 0$.
$\cos(x+\pi/3) \approx 
(1-x^2/2)/2 - x\sqrt{3}/2
\approx 1/2- x\sqrt{3}/2
$
and 
$\sin(3(x+\pi/3)) 
\approx 
-3x$.
Then $\sqrt{3+2\cos (x+\pi/3}-2 \approx
\sqrt{3+2(1/2- x\sqrt{3}/2)}
=\sqrt{4-x\sqrt{3}}
=2\sqrt{1-x\sqrt{3}/4}
\approx 2(1-x\sqrt{3}/8)
=2-x\sqrt{3}/4
$
so the numerator is about 
$-x\sqrt{3}/4$ as $x \to 0$.
For the denominator,
we have $\ln(1+x) \approx x$ as $x \to 0$,
so 
$\ln(1+\sin(3(x+\pi/3))
\approx \ln(1-3x)
\approx -3x$
so the ratio is
$\frac{-x\sqrt{3}/4}{-3x}
=\frac{\sqrt{3}}{12}
$.
This can be made rigorous if we can use
$\sin(x) = x +o(x^2)$
or
$\sin(x) \approx x +O(x^3)$
as $x \to 0$.
