TITLE: Finding a limit without series expansion and l'Hopital's rule: $\lim_{x\to 0}\frac{\tan x-\sin x}{\sin^3x}$
QUESTION [5 upvotes]: 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?

REPLY [7 votes]: $\frac{\tan x - \sin x}{(\sin x)^3} = \frac{\sin x (\frac{1}{\cos x}-1)}{(\sin x)^3} = \frac{\frac{1}{\cos x}-1}{1-(\cos x)^2} = \frac{1}{\cos x} \frac{1-\cos x}{(1-\cos x)(1+\cos x)} = \frac{1}{\cos x(1+\cos x)}$