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 :)
 A: Short answer: function discussion.
Long answer (details) is here.
First step. 
Your fraction is an even function so it is enough to consider the right limit.
Second step.
Define the functions $f(x):=\sin(x)-x+\frac{1}{6}x^3$ and $g(x):=\sin(x)-x+\frac{1}{6}x^3-\frac{1}{120}x^5$ on the interval $[0, 0.1]$.
Third step. Prove that $f$ is strictly increasing and $g$ is strictly decreasing. ($f(0)=0$, it is enough $f'>0$. $f'(0)=0$, so it is enough $f''>0$ and so on. Similarly for $g$. 
Forth step. From the third step 
$$
x-\frac{1}{6}x^3<\sin(x)<x-\frac{1}{6}x^3+\frac{1}{120}x^5.
$$
The limit follows.
A: Assuming $f$ is sufficiently smooth, repeated application of the fundamental theorem of the calculus gives (finite Taylor expansion)
$$f(x) = f(0)+f'(0)x+\frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \int_0^x (x-t)^2 \frac{(f'''(t)-f'''(0))}{2!}\, dt$$
Using the fact that $\sin' = \cos, \cos' = -\sin$, we can then expand $f(x) = \sin x$ as
$$ \sin x = x -\frac{x^3}{6} + \int_0^x (x-t)^2 \frac{(-\cos t +1)}{2!}\, dt$$
Let $\epsilon>0$, and choose $\delta>0$ such that if $|t|< \delta$, then $|1-\cos t| < \epsilon$. Then, replacing $1-\cos t$ by $\epsilon$ and integrating, we have the estimate
$$ \left|\sin x - x + \frac{x^3}{6} \right|  \leq \epsilon \frac{|x|^3}{6}$$
If $0 < |x| < \delta$, then dividing through by $|x^3|$ gives:
$$ \left| \frac{\sin x - x}{x^3} +\frac{1}{6} \right|  \leq \frac{\epsilon}{6} < \epsilon$$
The desired limit follows.
A: Let $L = \lim_{x \to 0} \dfrac{x - \sin(x)}{x^3}$. We then have
\begin{align}
L & = \underbrace{\lim_{y \to 0} \dfrac{3y - \sin(3y)}{27y^3} = \lim_{y \to 0} \dfrac{3y - 3\sin(y) + 4 \sin^3(y)}{27y^3}}_{\sin(3y) = 3 \sin(y) - 4 \sin^3(y)}\\
& = \lim_{y \to 0} \dfrac{3y - 3\sin(y)}{27 y^3} + \dfrac4{27} \lim_{y \to 0} \dfrac{\sin^3(y)}{y^3} = \dfrac{3}{27} L + \dfrac4{27}
\end{align}
This gives us $24L = 4 \implies L = \dfrac16$
A: 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
$$
\lim_{x\to0}\frac{x-\sin(x)}{x^3}=\frac16\tag5
$$
A: 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*}
A: Note: The original question didn't say anything about not using Taylor series.  Then, after I answered this, the OP changed the question and said didn't want Taylor series either.
Another option would be using power series (aka Maclaurin series aka Taylor series centered at 0)
$$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots$$
At this point, it's pretty easy.
