I have to find
$\lim_{x\to 0}\frac{\tan x-\sin x}{\sin^3x}$
without series expansion nor l'Hopital's rule, and am utterly and completely lost.
I ended up putting $x = 2y$ and getting to
$\lim_{y\to 0}\frac{1-\cos y-2\sin^2y\cos y}{\cos y\left(1-2\sin^2y\right)}\;,$
which gives me $0/1 = 0$, but the back of the book says that the answer is $1/2$.
How would this be solved without the use of series expansions nor l'Hopital's rule?