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.
 A: $$\lim_{x\to0}\dfrac{\cos3x-1}{\sin2x\tan3x}=-\lim_{x\to0}\dfrac{\sin^23x\cos3x}{\sin2x\sin3x(1+\cos3x)}$$
$$=-\dfrac32\cdot\lim_{x\to0}\dfrac{\sin3x}{3x}\cdot\dfrac1{\lim_{x\to0}\dfrac{\sin2x}{2x}}\cdot\lim_{x\to0}\dfrac{\cos3x}{1+\cos3x}=?$$
A: You have to recall three basic limits:
\begin{align}
&\lim_{t\to0}\frac{\sin t}{t}=1 \\[6px]
&\lim_{t\to0}\frac{\tan t}{t}=1 \\[6px]
&\lim_{t\to0}\frac{1-\cos t}{t^2}=\frac{1}{2}
\end{align}
In your problem, the coefficients of $x$ can be dealt with easily:
$$
\frac{ \cos (3x) - 1 }{ \sin (2x) \tan (3x)}
=
\frac{ \cos (3x) - 1 }{(3x)^2}\frac{2x}{\sin(2x)}\frac{3x}{\tan(3x)}
\frac{3^2}{2\cdot 3}
$$
where the last (numeric) fraction has been inserted to keep the expression the same.
Now compute the limit of each fraction and you're done.
A: \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}
A: $$\lim_{x\rightarrow0}\frac{\cos 3x - 1}{ \sin 2x \tan 3x }=-\lim_{x\rightarrow0}\frac{2\sin^2\frac{3x}{2}}{\sin2x\sin3x}=$$
$$=-\frac{3}{4}\lim_{x\rightarrow0}\frac{\frac{\sin^2\frac{3x}{2}}{\frac{9x^2}{4}}}{\frac{\sin2x}{2x}\cdot\frac{\sin3x}{3x}}=-\frac{3}{4}$$
A: 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 $\frac{37 \pi ^2}{2304}-\frac{3}{4}\approx -0.591504$.
A: 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.
