<s>[INST] Determine $\lim_{x \to 0}{\frac{x-\sin{x}}{x^3}}=\frac{1}{6}$, without L'Hospital or Taylor How can I prove that $$\lim_{x \to 0}{\frac{x-\sin{x}}{x^3}}=\frac{1}{6}$$
without using L'Hospital or Taylor series? 
thanks :)
 [/INST] First, an identity. For $x\ne0$,
$$
\begin{align}
\frac{2\sin(x)-\sin(2x)}{x^3}
&=\frac{2\sin(x)-2\sin(x)\cos(x)}{x^3}\tag{1a}\\
&=2\frac{\sin(x)}x\frac{1-\cos(x)}{x^2}\tag{1b}\\
&=\frac2{1+\cos(x)}\frac{\sin^3(x)}{x^3}\tag{1c}
\end{align}
$$
Explanation:
$\text{(1a)}$: $\sin(2x)=2\sin(x)\cos(x)$
$\text{(1b)}$: algebra
$\text{(1c)}$: $1-\cos(x)=\frac{\sin^2(x)}{1+\cos(x)}$
In this answer, it is shown geometrically that $\lim\limits_{x\to0}\frac{\sin(x)}x=1$. Thus, given $\epsilon\gt0$, we can choose $\delta\gt0$ so that if $0\lt|x|\le\delta$, then
$$
\frac2{1+\cos(x)}\frac{\sin^3(x)}{x^3}=[1-\epsilon,1+\epsilon]_\#\tag2
$$
where $[a,b]_\#$ is a real number in the interval $[a,b]$.
For $0\lt|x|\le\delta$, apply $(1)$ and $(2)$ to $x/2^{k+1}$ and divide by $2^{2k+3}$:
$$
\frac{2^{k+1}\sin\left(x/2^{k+1}\right)-2^k\sin\left(x/2^k\right)}{x^3}=\frac{[1-\epsilon,1+\epsilon]_\#}{2^{2k+3}}\tag3
$$
$\lim\limits_{x\to0}\frac{\sin(x)}x=1$ implies $\lim\limits_{k\to\infty}2^k\sin\left(x/2^k\right)=x$. Thus, if we sum $(3)$ in $k$, we get that for $0\lt|x|\le\delta$,
$$
\frac{x-\sin(x)}{x^3}=\frac{[1-\epsilon,1+\epsilon]_\#}6\tag4
$$
Therefore, the Squeeze Theorem yields
$$
\lim_{x\to0}\frac{x-\sin(x)}{x^3}=\frac16\tag5
$$
</s>