I'm wondering why the following use of little o is incorrect.
$$ \sin(x) = x - \frac{x^3}{6} + o(x^5) $$
I know that te definition of little o is the following:
$$ f(x) = o(g(x)) \quad \text{when } x \to x_0 \qquad \text{if } f(x)=g(x) \, w(x) \quad \text{with } \lim_{x \to x_0} w(x) = 0 $$
In my case should be $x_0=0$, $f(x) = x - \frac{x^3}{6}$, and $g(x) = x^3$. It seems that the limit tends to infinity, but the same situation happens if I consider (the following is correct):
$$ \sin(x) = x - \frac{x^3}{6} + o(x^4) $$
Answer suggested by the guys below: in the first case we have:
$$ w(x) = \frac{\sin(x) - \left( x - \frac{x^3}{3!} \right)}{x^5} = \frac{x - \frac{x^3}{3!} + \frac{x^5}{5!} - \left( x - \frac{x^3}{3!} \right)}{x^5} = \frac{1}{5!} $$
so:
$$ \lim_{x \to 0} w(x) = \frac{1}{5!} \neq 0 $$
Instead in the second case:
$$ w(x) = \frac{\sin(x) - \left( x - \frac{x^3}{3!} \right)}{x^4} = \frac{x - \frac{x^3}{3!} - \left( x - \frac{x^3}{3!} \right)}{x^4} = 0 $$
so:
$$ \lim_{x \to 0} w(x) = 0 $$
Your problem is that the sentence $\sin(x) = x - \frac{x^3}{6} + o(x^5)$ is NOT saying that $f(x) = o(x^5)$ where $f(x) = x - \frac{x^3}{6}$. It is saying that $f(x) = o(x^5)$ where $f(x) = \sin(x) - x + \frac{x^3}{6}$. Now use the Taylor expansion for sine.