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.
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}
&amp;=\frac{x-\frac{x^3}6+o(x^3)-x}{x^3}\\
&amp;=\frac{-\frac{x^3}6+o(x^3)}{x^3}\\
&amp;=-\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.
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