calculus - Find limit without using l'Hospital rule $limlimits_{xrightarrow0}frac{tan x-sin x}{x^3}$

How to find the following limit without using l'Hospital rule
$$\lim_{x\rightarrow0}\frac{\tan x-\sin x}{x^3}$$

Using l'Hospital I got $1\over2$. Thanks for your help.

Answer

Hints:

$$\frac{\tan x-\sin x}{x^3}=\frac{\sin x-\sin x\cos x}{x^3\cos x}=\frac1{\cos x}\frac{\sin x}x\frac{1-\cos x}{x^2}$$

Now, use arithmetic of limits and also

$$\frac{1-\cos x}{x^2}=\frac{\sin^2x}{x^2(1+\cos x)}=\left(\frac{\sin x}x\right)^2\frac1{1+\cos x}$$