<s>[INST] 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.
 [/INST] 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 $x \to 0$.
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