TITLE: Determine $\lim_{x \to 0}{\frac{x-\sin{x}}{x^3}}=\frac{1}{6}$, without L'Hospital or Taylor
QUESTION [11 upvotes]: 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 :)

REPLY [2 votes]: Reproduced from this answer, in which this answer is cited to show that $\lim\limits_{x\to0}\frac{\sin(x)}x=1$ and that $0\le x\lt\frac\pi2\implies0\le\sin(x)\le x\le\tan(x)$.

Pre-Calculus Proof that $\boldsymbol{\lim\limits_{x\to0}\frac{x-\sin(x)}{x^3}=\frac16}$
Assume that $0\lt x\le\frac\pi3$. Then, $\cos(x)\ge\frac12$ and $0\le\sin(x)\le x\le\tan(x)$. Therefore,
$$
\begin{align}
\frac{x-\sin(x)}{x^3}
&\le\frac{\tan(x)-\sin(x)}{x^3}\tag{1a}\\
&=\frac{\tan(x)}{x}\frac{1-\cos(x)}{x^2}\tag{1b}\\
&=\frac1{\cos(x)}\frac{\sin(x)}{x}\frac{2\sin^2(x/2)}{4\,(x/2)^2}\tag{1c}\\[6pt]
&\le1\tag{1d}
\end{align}
$$
Furthermore,
$$
\begin{align}
&\frac{x-\sin(x)}{x^3}-\frac14\frac{x/2-\sin(x/2)}{(x/2)^3}\tag{2a}\\
&=\frac{2(x/2)-2\sin(x/2)\cos(x/2)}{8(x/2)^3}-\frac{2(x/2)-2\sin(x/2)}{8(x/2)^3}\tag{2b}\\
&=\frac{2\sin(x/2)(1-\cos(x/2))}{8(x/2)^3}\tag{2c}\\
&=\frac{2\sin(x/2)\,2\sin^2(x/4)}{8(x/2)^3}\tag{2d}
\end{align}
$$
Since $\lim\limits_{x\to0}\frac{\sin(x)}x=1$, $(2)$ shows that
$$
\lim_{x\to0}\left(\frac{x-\sin(x)}{x^3}-\frac14\frac{x/2-\sin(x/2)}{(x/2)^3}\right)=\frac18\tag3
$$
For any $n$, adding $\frac1{4^k}$ times $(3)$ with $x\mapsto x/2^k$ for $k$ from $0$ to $n-1$  gives
$$
\begin{align}
\lim_{x\to0}\left(\frac{x-\sin(x)}{x^3}-\frac1{4^n}\frac{x/2^n-\sin\left(x/2^n\right)}{\left(x/2^n\right)^3}\right)
&=\frac18\frac{1-(1/4)^n}{1-1/4}\tag{4a}\\
&=\frac16-\frac16\frac1{4^n}\tag{4b}
\end{align}
$$
Thus, for any $\epsilon\gt0$, choose $n$ large enough so that $\frac1{4^n}\le\frac\epsilon2$. Then, $(4)$ says that we can choose a $\delta\gt0$ so that if $0\lt x\le\delta$,
$$
\frac{x-\sin(x)}{x^3}-\overbrace{\frac1{4^n}\frac{x/2^n-\sin\left(x/2^n\right)}{\left(x/2^n\right)^3}}^{\frac12[0,\epsilon]_\#}
=\frac16-\!\overbrace{\ \ \ \frac16\frac1{4^n}\ \ \ }^{\frac1{12}[0,\epsilon]_\#}\!+\frac12[-\epsilon,\epsilon]_\#\tag5
$$
where $[a,b]_\#$ represents a number between $a$ and $b$. The bounds above the braces follow from $(1)$ and the choice of $n$.
Equation $(5)$ says that for $0\lt x\le\delta$,
$$
\frac{x-\sin(x)}{x^3}=\frac16+[-\epsilon,\epsilon]_\#\tag6
$$
Since $\frac{x-\sin(x)}{x^3}$ is even, we can say that $(6)$ is true for $0\lt|x|\le\delta$, which means that
$$
\lim_{x\to0}\frac{x-\sin(x)}{x^3}=\frac16\tag7
$$