algebra precalculus - Finding a limit without series expansion and l'Hopital's rule: $lim_{xto 0}frac{tan x-sin x}{sin^3x}$

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?

Answer

Start with $\tan x=\frac{\sin x}{\cos x}$. With a little simplification our expression becomes

$$\frac{\sin x-\sin x\cos x}{\cos x\sin^3 x}.$$
Cancel a $\sin x$. We end up with
$$\frac{1-\cos x}{\cos x\sin^2 x}.$$
Now for the key trick: multiply top and bottom by $1+\cos x$. The top becomes $1-\cos^2 x$. This is $\sin^2 x$ and cancels the $\sin^2 x$ at the bottom. We end up with
$$\frac{1}{\cos x(1+\cos x)}.$$
Now finding the limit is straightforward, since $\cos x\to 1$ as $x\to 0$.