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

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?

• Please check to be sure that I didn’t introduce any errors when I converted to $\LaTeX$. Nov 21, 2012 at 1:37

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 $x\to 0$.
$\frac{\tan x - \sin x}{(\sin x)^3} = \frac{\sin x (\frac{1}{\cos x}-1)}{(\sin x)^3} = \frac{\frac{1}{\cos x}-1}{1-(\cos x)^2} = \frac{1}{\cos x} \frac{1-\cos x}{(1-\cos x)(1+\cos x)} = \frac{1}{\cos x(1+\cos x)}$