Trig limit without L'Hospital Rule: $\lim\limits_{x\to0}\frac{\tan x-\sin x}{x^3}$ I'm really getting stuck on this and would appreciate some help: 
$$
\lim_{x\ \to\ 0}\left[\,\tan\left(\,x\,\right) - \sin\left(\,x\,\right) \over x^{3}\,\right] 
$$
I know I need to change $\tan\left(\,x\,\right)$ into $\sin\left(\,x\,\right)/\cos\left(\,x\,\right)$  and turn $\sin\left(\,x\,\right)$ into $1 - \cos^{2}\left(\,x\,\right)$.
But then I get stuck. 
 A: $$\begin{align}\lim_{x\to 0}\frac{\tan x-\sin x}{x^3}&=\lim_{x\to 0}\frac{\sin x-\sin x\cos x}{x^3\cos x}\\&=\lim_{x\to 0}\frac{\sin x}{x}\cdot \frac{1-\cos x}{x^2}\cdot \frac{1}{\cos x}\\&=\lim_{x\to 0}\frac{\sin x}{x}\cdot\frac{\sin^2x}{x^2(1+\cos x)}\cdot\frac{1}{\cos x}\\&=\lim_{x\to 0}\frac{\sin x}{x}\cdot \left(\frac{\sin x}{x}\right)^2\cdot\frac{1}{1+\cos x}\cdot\frac{1}{\cos x}\\&=1\cdot 1^2\cdot \frac 12\cdot \frac{1}{1}\\&=\frac 12.\end{align}$$
A: Write the taylor series for the $\tan (x)$ and $\sin (x)$ after that solve the limit.
$$\lim_{x\ \to\ 0}\left[\,x+\frac {x^3}{3} - (x-\frac {x^3}{6} ) \over x^{3}\,\right] = \frac {1}{2}$$
A: Since $\lim_{x\to 0}\cos x=1$, we have:
$$ L=\lim_{x\to 0}\frac{\tan x-\sin x}{x^3}=\lim_{x\to 0}\frac{\sin x-\frac{1}{2}\sin(2x)}{x^3},$$
and since $\sin x = x-\frac{x^3}{6}+o(x^4)$ in a neighbourhood of zero, 
$$ L = -\frac{1}{6}+\frac{1}{2}\cdot\frac{8}{6}=\color{red}{\frac{1}{2}}.$$
A: Perhaps overkill. Remember that
$$
\sin x=\frac{2\tan(x/2)}{1+\tan^2(x/2)},\qquad
\tan x=\frac{2\tan(x/2)}{1-\tan^2(x/2)}
$$
so
$$
\tan x-\sin x=
2\tan\frac{x}{2}\frac{1+\tan^2(x/2)-1+\tan^2(x/2)}{1-\tan^4(x/2)}
=\frac{4\tan^3(x/2)}{1-\tan^4(x/2)}
$$
so your limit is
$$
\lim_{x \to 0}\frac{\tan x - \sin x}{x^{3}}=
\lim_{x\to 0}\frac{4\tan^3(x/2)}{8(x/2)^3}\frac{1}{1-\tan^4(x/2)}=
\frac{4}{8}=\frac{1}{2}
$$
since $\lim_{t\to0}\dfrac{\tan t}{t}=1$.
