Tangent minus sine over x cubed

Idea. Both $\tan x$ and $\sin x$ behave like $x$, so their difference is much smaller than $x$. Factoring reveals the two basic limits we know.

Computation. Since $\sin x=\tan x\cos x$,
$$\tan x-\sin x=\tan x\,(1-\cos x),$$
and therefore
$$\frac{\tan x-\sin x}{x^3}=\frac{\tan x}{x}\cdot\frac{1-\cos x}{x^2}.$$
We know that $$\dfrac{\tan x}{x}=\dfrac{\sin x}{x}\cdot\dfrac1{\cos x}\to1\cdot1=1$$ and $$\dfrac{1-\cos x}{x^2}\to\dfrac12.$$ The limit is $\dfrac12$.

Remark. Splitting a quotient into a product of known limits works because the limit of a product is the product of the limits, as long as each factor has a limit.