Calculate: $\lim_{x\to 0}{\frac{\tan x-\sin x}{x^3}}$ 
Calculate: $\lim{\frac{\tan x-\sin x}{x^3}}$ as $x\to 0$   

I have solved it using a way that gives a wrong answer but I can't figure out why:
$$\lim{\frac{\tan x-\sin x}{x^3}}\\
=\lim{\frac{\tan x}{x^3}-\lim\frac{\sin x}{x^3}}\\
=\lim{\frac{1}{x^2}}\cdot\lim{\frac{\tan x}{x}-\frac{1}{x^2}}\cdot\lim{\frac{\sin x}{x}}\\
=\lim{\frac{1}{x^2}}-\lim{\frac{1}{x^2}}
=0$$
The answer using the standard method gives $\frac{1}{2}$.
It would be great if someone could explain to me why this method is wrong. 

$\frac{\tan{x}-\sin{x}}{x^3}=\frac{2\sin{x}\sin^2\frac{x}{2}}{x^3\cos{x}}\rightarrow2\cdot\left(\frac{1}{2}\right)^2=\frac{1}{2}$
