<s>[INST] calculating a limit using taylor series $$\lim_{x\rightarrow0}\frac{\cos(x) \sin(x)-x}{\sin^3 x}$$
I tried to use the identity $\cos(x) \sin(x)=\frac{1}{2}\sin(2x)$
and then use the taylor $\lim_{x\rightarrow0}\frac{x+\alpha(x)}{x^3+\beta(x)}$ where $\alpha(x), \beta(x)$ are the remainders which leads me to infinity but it's not the right answer because I did the same with l'hopital rule and got $-\frac{3}{2}$ (I used it three times!) 
Can anyone help me how to solve it using taylor series?
Thanks in advance!
 [/INST] $$\lim _{x\rightarrow 0} \frac {\cos x \sin x -x }{\sin^3 x} =\lim _{x\rightarrow 0} \frac { \sin 2x -2x }{2\sin^3 x} =\lim_{x\rightarrow 0} \frac { 2x-\frac {8x^3}{3!} +o(x^3) -2x }{ 2\sin^3 x} =\\
=\lim_{x\rightarrow 0} \frac {-\frac {8x^3}{3!} +o(x^3)}{2\sin^3 x} =\lim_{x\rightarrow 0} -\frac {8x^3}{12\sin^3 x} +\lim _{x\rightarrow 0} \frac { o\left(x^3 \right)}{2\sin^3 x} =-\frac 2 3$$
And by L'Hospital's rule $$\lim _{x\rightarrow 0} \frac {\cos x \sin x -x }{\sin^3 x} =\lim_{x\rightarrow 0} \frac {\sin 2x -2x}{2\sin^3 x} \overset {\text{L'Hospital}} = \lim_{x\rightarrow 0} \frac{2\cos 2x -2 }{6\sin^2 x\cos x} =\\
=4\lim_{x\rightarrow 0} \frac {-\sin^2 x}{6\sin^2 x\cos x} =-\frac 2 3$$
</s>