<s>[INST] Find the limit of $ \frac{\cos (3x) - 1}{ \sin (2x) \tan (3x) } $ without L'Hospital technique I would like to find the limit as $x$ goes to zero for the following function, but without L'Hospital technique.
$$ f(x) = \frac{ \cos (3x) - 1 }{ \sin (2x) \tan (3x)}   $$
This limit will go to zero (I had tried using calculator manually).
I have tried to open the trigonometric identities and the fact that $ \lim\limits_{x \rightarrow 0}\frac{\sin (ax)}{x} = a $ gives :
$$  \lim_{x \rightarrow 0} f(x) = \lim_{x \rightarrow 0} \frac{\cos (3x)( \cos (3x) - 1)}{6 x^{2}}  $$
$$ \cos (3x) = \cos (2x) \cos(x) - \sin (2x) \sin(x) = \cos^{3}(x) - 3\sin^{2}(x) \cos(x)  $$
$$ \lim_{x \rightarrow 0 } \frac{ \cos (3x) }{x^{2}} = \lim_{x \rightarrow 0} \frac{\cos^{3}(x)}{x^{2}} - 3  $$
Any view on this and more efficient way to solve this? Thanks before.
 [/INST] \begin{align}
\lim_{x\to0} \frac{ \cos (3x) - 1 }{ \sin (2x) \tan (3x)} 
&= \lim_{x\to0} \frac{ -2\sin^2 (\dfrac{3x}{2}) \cos (3x) }{ \sin (2x) \sin (3x)} \\
&=  -2\lim_{x\to0} \dfrac{\sin^2 (\dfrac{3x}{2}) }{(\dfrac{3x}{2})^2} \dfrac{(2x)}{\sin (2x)} \frac{ (3x)}{ ( \sin (3x)}.\dfrac{(\dfrac{3x}{2})^2}{(2x)(3x)}\cos (3x) \\
&=\color{blue}{-\dfrac{3}{4}}
\end{align}
</s>