Calculating $\lim_{x \rightarrow 0} \frac{\tan x - \sin x}{x^3}$. I have a difficulty in calculating this limit: 
$$\lim_{x \rightarrow 0} \frac{\tan x - \sin x}{x^3},$$
I have tried $\tan x = \frac{\sin x}{\cos x}$, then I unified the denominator of the numerator of the given limit problem finally I got $$\lim_{x \rightarrow 0} \frac{\sin x}{x^{3} \cos x} - \lim_{x \rightarrow 0} \frac{ \sin x}{x^3},$$
Then I got stucked, could anyone help me in solving it?   

$$\frac{\tan{x}-\sin{x}}{x^3}=\frac{x+\frac{1}{3}x^3+o(x^3) - (x-\frac{1}{6}x^3+o(x^3))}{x^3}=\frac{\frac{1}{2}x^3+o(x^3)}{x^3}=\frac{1}{2} +\frac{o(x^3)}{x^3}\stackrel{x\rightarrow 0}{\longrightarrow}\frac{1}{2}$$
