I'm familiar to both the $O$ and little $o$ notation. I know they are of great use when studying limits, asymptotics and series. However, I'm not sure when to use one over another, particularly when solving problems. For example, you can put that, for $x \to 0$
$\sin x = x + o(x)$
since
$\frac{\sin x}{x}-1 \to 0$
I've also recently read about a great definition for the derivative and strong derivative (Knuth), respectively, for $\epsilon \to 0$:
f(x+\epsilon)=f(x)+\epsilon f'(x)+o(\epsilon)
which clearly means that
\frac{f(x+\epsilon)-f(x)}{\epsilon}- f'(x) \to 0
and the same for
f(x+\epsilon)=f(x)+\epsilon f'(x)+o(\epsilon ^2)
Now consider the following
$\lim_{x \rightarrow 0}{\frac{\int_{0}^{x} \frac{\sin(t)}{t}-\tan(x)}{2x(1-\cos(x))}}$
I know that
$2x\left( {1 - \cos x} \right) = {x^3} + o\left( {{x^3}} \right)$
$\tan x = x + \frac{x^3}{3} + o\left( x^3 \right)$
$\int\limits_0^x {\frac{{\sin t}}{t}dt = x -\frac{x^3}{3\cdot 3!} +o\left( x^3 \right)} $
So this gives
$\frac{{x - \frac{{{x^3}}}{{3 \cdot 3!}} + o\left( {{x^3}} \right) - x - \frac{{{x^3}}}{3} - o\left( {{x^3}} \right)}}{{{x^3} + o\left( {{x^3}} \right)}} = $
$ \frac{{\left( { - \frac{1}{3} - \frac{1}{{18}}} \right){x^3} + o\left( {{x^3}} \right)}}{{{x^3} + o\left( {{x^3}} \right)}}$
$\frac{{\left( { - \frac{7}{{18}}} \right){x^3} + o\left( {{x^3}} \right)}}{{{x^3} + o\left( {{x^3}} \right)}} = - \frac{7}{{18}}$
So how could I use the big $O$ to solve the same problem? And why should I choose big $O$ over little $o$, when $o$ seems more accurate than $O$?
NOTE: When I talk about the big $O$ I'm considering it for any $x \to a$, not only for asymptotic behaviour $x \to \infty$. For example, for $x \to 0$
$\sin x = O(x) $