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 
$$
  \lim_{ x\to 0 } \frac{x\cos(x) - \sin(x)}{2  x^3} = \frac{-1}{6}
$$

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
$$\lim_{x\to 0}\dfrac{x\cos{x}-\sin{x}}{2x^3}=-\dfrac{1}{6}$$
