Problems with limits and asymptotic notation I have trouble understanding when it's ok or not to substitute a function with another asymptotic one.
I know that, when $\lim _{x\to x_0}\left(\frac{f\left(x\right)}{g\left(x\right)}\right)=1$ , if I'm in a limit in which $x\to x_0$ , then I can "replace" f(x) with g(x) .
But WHEN can I replace it?
An example that is making me confused:
Consider this limit:  
$\lim _{x\to \:0}\left(\frac{sin\left(x\right)-x+2x^5}{3x^3}\right)$
The result should be $\ell =-\frac{1}{18}$ (Wolfram agrees)
But I'm having trouble with this.
First error I noticed was trying to apply the notable limit $\lim _{x\to \:0}\left(\frac{sin\left(x\right)}{x}\right)=1$.
I grouped by x, and I applied the substitution, and I ended with a wrong result ($\ell =0$).
$\lim \:_{x\to \:\:0}\left(\frac{sin\left(x\right)-x+2x^5}{3x^3}\right)=\lim \:_{x\to \:\:0}\left(\frac{x\left(\frac{sin\left(x\right)}{x}-1+\frac{2x^5}{x}\right)}{3x^3}\right)=\lim \:\:_{x\to \:\:\:0}\left(\frac{x\left(\left[1\right]-1+\frac{2x^5}{x}\right)}{3x^3}\right)=...=0$.
Ok, this is wrong.
But why THIS other approach I tried is wrong?  
$\lim \:_{x\to \:\:0}\left(\frac{sin\left(x\right)-x+2x^5}{3x^3}\right)$ ⇒ For x→0, sin⁡(x)∼x ⇒ $\lim \:_{x\to \:\:0}\left(\frac{\left[x\right]-x+2x^5}{3x^3}\right)=...=0$ 
WHEN is it not appropriate to substitute a function with an asymptot?
Help.
 A: You cannot sum this type of relation only multiply. If you wanna study your example you cannot suppress the "rest" behind it because he's too important in the considered limit :

$$
\frac{\sin\left(x\right)-x+2x^2}{3x^3}=\frac{\left(x-\frac{x^3}{6}-x+2x^5+o\left(x^3\right)\right)}{3x^3} \underset{x \rightarrow 0}{\rightarrow}\frac{1}{6 \times 3}=-\frac{1}{18}
$$

The thing is here you used $$\sin\left(x\right) \underset{\left(0^{+}\right)}{\sim}x$$ and you summed equivalent. You can substitute when it is a product, not a sum. For example you would have had instead
$$
\frac{2\sin\left(x\right)x^2}{3x^3}
$$
You could have wrote

$$
\frac{2\sin\left(x\right)x^2}{3x^3} \underset{\left(0^{+}\right)}{\sim}\frac{2xx^2}{3x^3}\underset{x \rightarrow 0}{\rightarrow}\frac{2}{3}
$$

A: This kind of asymptotic substituition is very complicated and Dangerous to be handled for limits, it is better to proceed by Taylor's expansion with little-o or big-O notation.
In this case note that
$$\sin x= x -\frac{x^3}{6}+o(x^3)$$
then
$$\frac{sin\left(x\right)-x+2x^5}{3x^3}=\frac{x -\frac{x^3}{6}+o(x^3)-x+2x^5}{3x^3}=\frac{-\frac{x^3}{6}+o(x^3)}{3x^3}=\frac{-\frac{1}{6}+o(1)}{3}\to-\frac1{18}$$
Thus in this case you can't replace $\sin x \sim x$ because you need higher order terms to obtain the correct limit. In Others case the first order expansion might be sufficient and in others case you will need to expand further. There are not general rules for it.
