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 :)

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.
