Problem with limit solving I've come across this limit:
$$\lim_{x\to 0}\frac{x\cos{x}-\sin{x}}{\sin^3{x}}$$
My idea was to split this up into two fractions to obtain:
$$\lim_{x\to 0}\bigg(\frac{x\cos{x}}{\sin^3{x}}-\frac{1}{\sin^2{x}}\bigg)$$
in the first fraction i use the fact that $x$ and $\sin{x}$ are simmilar near zero therefore the quotiont of these two is equal to $1$.
Moreover (getting back to common denominator)
$$\lim_{x\to 0}\frac{\cos{x}-1}{\sin^2{x}}=-\lim_{x\to 0}\frac{1-\cos{x}}{1-\cos^2{x}}=-\lim_{x\to 0}\frac{1}{1+\cos{x}}=-\frac{1}{2}$$
which IS supposed to be wrong When i apply L'Hospital's rule:
$$\lim_{x\to 0}\frac{x\cos{x}-\sin{x}}{\sin^3{x}}=\lim_{x\to 0}\frac{-x\sin{x}+\cos{x}-\cos{x}}{3\sin^2{x}\cos{x}}=\lim_{x\to 0}\frac{-x}{3\sin{x}}=-\frac{1}{3}$$
which is the correct answer. Would anyone give me an explanation, what exactly i did wrong there? And is there any different solution without LH which would get me to desired $-\frac{1}{3}$?
 A: Note that this step
$$\lim_{x\to 0}\bigg(\frac{x\cos{x}}{\sin^3{x}}-\frac{1}{\sin^2{x}}\bigg)=\lim_{x\to 0}\frac{\cos{x}-1}{\sin^2{x}}$$
is not allowed since you are solving a part of the limit and approximating $\frac{\sin x}{x}$ with $1$ that's not correct since you are neglecting higher order terms (indeed the correct solution requires to apply l'Hospital's rule twice).
To understand this point let consider the limit for $x\to 0$
$$\frac{\frac{\sin x}{x}-1}{x^2}\not \Rightarrow \frac{1-1}{x^2}=0$$
but 
$$\frac{\frac{\sin x}{x}-1}{x^2}=\frac{1-\frac{x^2}6-1+o(x^2)}{x^2}=-\frac16+o(1)\to -\frac16$$
Refer to this OP for a general explanation of this fact Analyzing limits problem Calculus (tell me where I'm wrong).
A: You have transformed the indeterminate form $0/0$ into $\infty-\infty$ and, no, you cannot use that $x\sim\sin x$.
You can rather rewrite the limit as
$$
\lim_{x\to0}\cos x\frac{x^3}{\sin^3x}\frac{x-\tan x}{x^3}
$$
and the first two factors will contribute $1$, provided the limit of the third factor exists. Note that removing those factors is possible, because they are factors, not summands.
Not really an advancement, to be honest, but useful to see where your method goes wrong. If you apply it and split the fraction into $1/x^2-\tan x/x^3$ (again $\infty-\infty$), observing that $\tan x\sim x$ would lead to $1/x^2-1/x^2=0$. You clearly see that this method cannot work: applying it to the same limit in two different forms yields different results.
I see no nice and simple way to compute the limit without resorting to l'Hôpital or Taylor. With Taylor:
\begin{align}
\lim_{x\to 0}\frac{x\cos x-\sin x}{\sin^3x}
&=\lim_{x\to0}\frac{x(1-x^2/2+o(x^2))-(x-x^3/6+o(x^3))}{(x+o(x))^3} \\[6px]
&=\lim_{x\to0}\frac{x^3(-1/2+1/6)+o(x^3)}{x^3+o(x^3)} \\[6px]
&=-\frac{1}{3}
\end{align}
Or, with l'Hôpital once,
$$
\lim_{x\to0}\frac{x\cos x-\sin x}{\sin^3x}=
\lim_{x\to0}\frac{\cos x-x\sin x-\cos x}{3\sin ^2x\cos x}=
\lim_{x\to0}\frac{-x}{3\sin x\cos x}=-\frac{1}{3}
$$
using $\sin x\sim x$.
