Trigonometric Limit without L'Hopital I am having problems solving this limit without L'Hopital or series.
$$
  \lim_{ x\to 0 } \frac{x\cos(x) - \sin(x)}{2  x^3}
$$
I tried some trigonometric manipulations without success. I tried Trigonometric identities with no luck and separating $$  \frac{x\cos(x)}{2  x^3}   and   \frac{sin(x)}{2  x^3} $$ lead me nowhere, each of this limits are infinity. I kow the result is 
$$
  \lim_{ x\to 0 } \frac{x\cos(x) - \sin(x)}{2  x^3} = \frac{-1}{6}
$$
 A: Write the original expression as follows
\begin{equation*}
\frac{x\cos x-\sin x}{2x^{3}}=\frac{x\cos x-x+x-\sin x}{2x^{3}}=\frac{1}{2}%
\left( \frac{\cos x-1}{x^{2}}+\frac{x-\sin x}{x^{3}}\right) .
\end{equation*}
Now use the standard limits
\begin{equation*}
\lim_{x\rightarrow 0}\frac{\cos x-1}{x^{2}}=-\frac{1}{2},\ \ \ \ \ \ and\ \
\ \ \ \ \ \ \lim_{x\rightarrow 0}\frac{x-\sin x}{x^{3}}=\frac{1}{6}
\end{equation*}
it follows
\begin{equation*}
\lim_{x\rightarrow 0}\frac{x\cos x-\sin x}{2x^{3}}=\frac{1}{2}\left( -\frac{1%
}{2}+\frac{1}{6}\right) =-\frac{1}{6}.
\end{equation*}
A: since
$$\lim_{x\to 0}\dfrac{x\cos{x}-\sin{x}}{2x^3}=\lim_{x\to 0}\dfrac{-\cos{x}(\tan{x}-x)}{2x^3}=-\lim_{x\to 0}\dfrac{\tan{x}-x}{2x^3}$$
Use this two inequality
$$\tan{x}>x+\dfrac{1}{3}x^3\Longrightarrow \dfrac{\tan{x}-x}{2x^3}>\dfrac{1}{6},x\in [0,\dfrac{\pi}{2})$$
Other hand
$$\tan{x}<\dfrac{3x}{3-x^2}\Longrightarrow \dfrac{\tan{x}-x}{2x^3}<\dfrac{1}{2(3-x^2)},x\in [0,\dfrac{\pi}{2})$$
so
$$\lim_{x\to\infty}\dfrac{\tan{x}-x}{2x^3}=\dfrac{1}{6}$$
so
$$\lim_{x\to 0}\dfrac{x\cos{x}-\sin{x}}{2x^3}=-\dfrac{1}{6}$$
