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.

The thinking/attempt in the question post is quite complicated. Simple idea related to @ParamanandSingh 's comment. Multiply the function by $ \frac{\cos(3x)+1}{\cos(3x)+1}  $, we get
$$ f(x) = \frac{\cos^{2}(3x) - 1}{\sin(2x) \tan(3x)(\cos(3x)+1)} =\frac{- \sin(3x) \cos(3x)}{\sin(2x) (\cos(3x)+1)} = \frac{- (2x) \sin(3x)  \cos(3x)}{(3x) \sin(2x) (\cos(3x)+1)} \left( \frac{3}{2} \right)   $$
As $x \rightarrow 0$, the limit will go to
$$  \frac{-1}{2} \left(\frac{3}{2} \right) = -\frac{3}{4} $$
Thanks for the inputs on this question.
