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 :)

The following argument is based on the suggestions by vesszabo.  (I do not restrict the functions to the interval $[0, \, 0.1]$ as vesszabo did. That would erroneous because we are looking for the limit at $0$.)  It is a rigorous argument and it does avoid using Taylor's Theorem ... but it is still hokey.  The choice to use the functions $f$ and $g$ is guided by Taylor's Theorem. 
Demonstration
$f$ is a function defined by
\begin{equation*}
f(x) = \sin{x} - x + \frac{1}{6} \, x^{3} .
\end{equation*}
This function is differentiable, and
\begin{equation*}
f^{\prime}(x) = \cos{x} - 1 + \frac{1}{2} \, x^{2} .
\end{equation*}
Likewise, this derivative is differentiable, and
\begin{equation*}
f^{\prime\prime}(x) = -\sin{x} + x .
\end{equation*}
$f(0) = f^{\prime}(0) = 0$, but since $f^{\prime\prime}(x) > 0$ for every positive, real number $x$, $f^{\prime}(x)$ is an increasing function, and since $f^{\prime}(0) = 0$, $f^{\prime}(x) > 0$ for every positive, real number $x$. According to a corollary to the Mean Value Theorem, $f$ is an increasing function. $f(0) = 0$, and so $f(x) > 0$ for every positive, real number $x$.
$g$ is another twice-differentiable function defined by
\begin{equation*}
g(x) = \sin{x} - x + \frac{1}{6} \, x^{3} - \frac{1}{120} x^{5} = f(x) - \frac{1}{120} x^{5} .
\end{equation*}
For every positive, real number $x$,
\begin{equation*}
g^{\prime\prime}(x)
= f^{\prime\prime}(x) - \frac{1}{6} x^{3}
= - f(x)
< 0 .
\end{equation*}
According to a corollary to the Mean Value Theorem, $g^{\prime}$ is a decreasing function. $g^{\prime}(0) = 0$, and so $g^{\prime}(x) < 0$ for every positive, real number $x$. According to a corollary to the Mean Value Theorem, $g$ is a decreasing function. $g(0) = 0$, and so $g(x) < 0$ for every positive, real number $x$.
So, for every positive, real number $x$,
\begin{equation*}
\frac{1}{6} \, x^{3} - \frac{1}{120} x^{5}
< x - \sin{x}
< \frac{1}{6} \, x^{3} ,
\end{equation*}
and
\begin{equation*}
\frac{1}{6} - \frac{1}{120} x^{2}
< \frac{x - \sin{x}}{x^{3}}
< \frac{1}{6} .
\end{equation*}
According to the Squeeze Theorem,
\begin{equation*}
\lim_{x\to0^{+}} \frac{x - \sin{x}}{x^{3}} = \frac{1}{6} .
\end{equation*}
$f$ and $g$ are odd functions. So, for every negative, real number $x$,
\begin{equation*}
\frac{1}{6} \, x^{3}
< x - \sin{x}
< \frac{1}{6} \, x^{3} - \frac{1}{120} x^{5} ,
\end{equation*}
and
\begin{equation*}
\frac{1}{6} - \frac{1}{120} x^{2}
< \frac{x - \sin{x}}{x^{3}}
< \frac{1}{6} .
\end{equation*}
Again, according to the Squeeze Theorem,
\begin{equation*}
\lim_{x\to0^{-}} \frac{x - \sin{x}}{x^{3}} = \frac{1}{6} .
\end{equation*}
A function that has the same left-sided and right-sided limits at $0$ has a limit at $0$.
\begin{equation*}
\lim_{x\to0} \frac{x - \sin{x}}{x^{3}} = \frac{1}{6} .
\end{equation*}
