$ \lim_{x\to 0 } \frac{\tan x - \sin x}{x^3}$ using L'Hopital $$\displaystyle \lim_{ x\to 0} \frac{\tan x - \sin x}{x^3}$$
$$ \displaystyle \lim_{ x\to 0} \frac{\sec^2x - \cos x}{3x^2}$$
$$ \displaystyle \lim_{x\to 0} \frac{2\cos^{-3}x \sin x + \sin x}{6x}$$
Is it indeed complicated using LHopital, how do I continue?

You can pull out a $\tan x$ factor and
$$\frac{\tan x-\sin x}{x^3}=\frac{\tan x}x\frac{1-\cos x}{x^2}=\frac{\tan x}x\frac{2\sin^2\dfrac x2}{x^2}.$$
This is enough to conclude
$$\to\frac12.$$

Direct L'Hospital is manageable
$$\frac{\tan x-\sin x}{x^3}\to\frac{\tan^2x+1-\cos x}{3x^2}\to\frac{2\tan x(\tan^2x+1)+\sin x}{6x}\to\frac{2+1}6,$$
but easier after pulling $\tan x$,
$$\frac{\tan x}x\frac{1-\cos x}{x^2}\to\frac{\tan x}x\frac{\sin x}{2x}.$$
