Trig limit without L'Hospital Rule: $\lim\limits_{x\to0}\frac{\tan x-\sin x}{x^3}$ I'm really getting stuck on this and would appreciate some help: 
$$
\lim_{x\ \to\ 0}\left[\,\tan\left(\,x\,\right) - \sin\left(\,x\,\right) \over x^{3}\,\right] 
$$
I know I need to change $\tan\left(\,x\,\right)$ into $\sin\left(\,x\,\right)/\cos\left(\,x\,\right)$  and turn $\sin\left(\,x\,\right)$ into $1 - \cos^{2}\left(\,x\,\right)$.
But then I get stuck. 

$$\begin{align}\lim_{x\to 0}\frac{\tan x-\sin x}{x^3}&=\lim_{x\to 0}\frac{\sin x-\sin x\cos x}{x^3\cos x}\\&=\lim_{x\to 0}\frac{\sin x}{x}\cdot \frac{1-\cos x}{x^2}\cdot \frac{1}{\cos x}\\&=\lim_{x\to 0}\frac{\sin x}{x}\cdot\frac{\sin^2x}{x^2(1+\cos x)}\cdot\frac{1}{\cos x}\\&=\lim_{x\to 0}\frac{\sin x}{x}\cdot \left(\frac{\sin x}{x}\right)^2\cdot\frac{1}{1+\cos x}\cdot\frac{1}{\cos x}\\&=1\cdot 1^2\cdot \frac 12\cdot \frac{1}{1}\\&=\frac 12.\end{align}$$
