<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] Just another way using the standard Taylor expansions.
$$\cos(3x)=1-\frac{9 x^2}{2}+\frac{27 x^4}{8}+O\left(x^6\right)$$
$$\sin(2x)=2 x-\frac{4 x^3}{3}+O\left(x^5\right)$$
$$\tan(3x)=3 x+9 x^3+O\left(x^5\right)$$
makes
$$\cos(3x)-1=-\frac{9 x^2}{2}+\frac{27 x^4}{8}+O\left(x^6\right)$$
$$\sin(2x)\tan(3x)=6 x^2+14 x^4+O\left(x^6\right)$$ Now, long division leads to 
$$\frac{\cos (3x) - 1}{ \sin (2x) \tan (3x) }=-\frac{3}{4}+\frac{37 x^2}{16}+O\left(x^3\right)$$ showing the limit and how it is appraoched.
Just for illustration, use $x=\frac \pi {12}$ (quite far away from $0$); the exact value of the expression is $\sqrt{2}-2\approx -0.585786$ while the above approximation would give $\frac{37 \pi ^2}{2304}-\frac{3}{4}\approx -0.591504$.
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