What is the real thing when using Taylor to prove limits, such as $\lim_{x \to 0}\frac{\sin x-x}{x^3}=\frac16$? I see some answers in this site proving many limits by Taylor, for example 
$$\lim_{x \to 0}\dfrac{\sin x-x}{x^3}=-\dfrac16$$ 
comes to my mind. However, my teacher didn't teach me this technique, also the popular calculus/analysis books I have do not even mention this technique. When people use Taylor to compute this limit, they often use big-$O$ operations, or omitting the terms having high order. However, what are the big-$O$ techniques? How does they work? Why is this rigorous? And how to choose where the omitting occurs? Should I use Taylor "expansions" (with Lagrange remainder) or Taylor "series"? I totally have no idea. Need your help.
 A: Formally, a Taylor expansion up to the $k$-th term means that
$$f(x+h)=f(x)+hf'(x)+\cdots+\frac 1{k!}h^k f^{(k)}(x)+\color{red}{o(h^k)}$$
where the little $o$-term is a placeholder for a not further determined function $g(h)$ with
$$\lim_{h\to 0}\frac{g(h)}{h^k}=0$$
Because this fraction vanishes for small $h$, it vanishes in limits as the one in your example. This is useful in situations where we have an approximation for a certain function which is not perfect, but we want to express that we know this and that we can say how good/bad it approximates.

Example. Choose $f(x)=\sin(x)$ and observe
$$\sin(x)=\sin(0)+x\cos(0)-\frac{x^2}2\sin(0)-\frac{x^3}6\cos(0)+o(x^3)=x-\frac{x^3}6+o(x^3).$$
Now consider your example:
\begin{align}
\frac{\sin(x)-x}{x^3}
&=\frac{x-\frac{x^3}6+o(x^3)-x}{x^3}\\
&=\frac{-\frac{x^3}6+o(x^3)}{x^3}\\
&=-\frac16+\frac{o(x^3)}{x^3}\\
\end{align}
and we know that $o(x^3)/x^3\to0$ for $x\to 0$. So the limit is $-1/6$. See that we do not actually need to know what the function $o(x)$ stands for, as we only need that it vanishes in certain limits. This is the power of the Landau $\mathcal O$-notation (I could have used $\mathcal O(x^{4})$ instead, but I personally find $o(x^3)$ more appealing here). Of course the really powerful result here is Taylor's theorem which provides you with this $o(x^3)$ term in the first place.
A: That is quite a few questions.
$O$-notation:
A function $f(x) = O(g(x))$ as $x \to x_0$ if there exist a neighbourhood $B(x_0,\epsilon)$ of $x_0$ and $M>0$ such that $|f(x)| \leq M |g(x)|$, for $x \in B(x_0,\epsilon)$.
Connection to taylors formula with remainder:
Using taylors formula with remainder we have 
$$
f(x) = \sum_{k=0}^{n-1} \frac{f^{(k)}(x_0)}{k!}(x-x_0)^k + \frac{f^{(n)}(z)}{n!}(x-x_0)^n,
$$
where $z$ lies between $x$ and $x_0$. If $f^{(n)}$ is continuous near $x_0$ we have that $|f^{(n)}(z)| \leq \max_{z \in [x_0- \epsilon, x_0+\epsilon]}|f^{(n)}(z) |=M_1$. Setting $M=M_1/n!$ we get that:
$$
|f(x) - \sum_{k=0}^{n-1} \frac{f^{(k)}(x_0)}{k!}(x-x_0)^k| \leq M|(x-x_0)^n|
$$
near $x_0$. Therefore $f(x)-\sum_{k=0}^{n-1} \frac{f^{(k)}(x_0)}{k!}(x-x_0)^k = O(x-x_0)^n$ as $x\to x_0$.
A little calculus with $O$-notation: Assume $f(x) = O((x-x_0)^n)$ as $x\to x_0$ and $0<k \leq n $ then $f(x)/(x-x_0)^k = O((x-x_0)^{n-k})$. This follows easily from the fact that in $B(x_0,\epsilon)$ we have that:
$$
 |\frac{f(x)}{(x-x_0)^k}| \leq \frac{M|(x-x_0)^n|}{|(x-x_0)^k|} = M|(x-x_0)^{n-k}|.
$$
Now for your problem:
We know from taylor (and the above connection with $O$-notation) that $\sin(x) = x-\frac16 x^3 +O(x^4)$ as $x\to 0$. Now we have that $\frac{\sin(x) -x}{x^3} -(- \frac16) = O(x)$. That is $|\frac{\sin(x) -x}{x^3} - (-\frac16)| \leq M |x| $ as $x \in B(0,\epsilon)$. Thus $\lim_{x\to 0}\frac{\sin(x) -x}{x^3}=-\frac16$.
(Btw: $f(x)=g(x)+O(h(x)$ means that $f(x)-g(x) = O(h(x))$)
A: Hint: $\sin(x)=x-x^3/6+ O(x^4)$ This implies that ${{\sin(x)-x}\over x^3}=-1/6+O(x)$.
https://en.wikipedia.org/wiki/Taylor_series#Trigonometric_functions
