# Why $\sin(x) = x - \frac{x^3}{6} + o(x^5)$ is false?

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:

$$\lim_{x \to 0} w(x) = \frac{1}{5!} \neq 0$$

$$w(x) = \frac{\sin(x) - \left( x - \frac{x^3}{3!} \right)}{x^4} = \frac{x - \frac{x^3}{3!} - \left( x - \frac{x^3}{3!} \right)}{x^4} = 0$$

so:

$$\lim_{x \to 0} w(x) = 0$$

• Consider $\lim_{x\to0}\frac{\sin x -(x-\frac{x^3}6)}{x^5}$.
– user
Feb 18, 2022 at 13:15
• @user thank you very much for your useful comment. Feb 24, 2022 at 9:02
• $\frac{\sin(x) - \left( x - \frac{x^3}{3!} \right)}{x^4} = \frac{x - \frac{x^3}{3!} - \left( x - \frac{x^3}{3!} \right)}{x^4}$ is not true since $\sin x \neq x - \frac{x^3}{3!}$.
– Gary
Feb 24, 2022 at 10:02
• Hello @Gary can you write your answer please? It'd be easier to understand the right result Feb 24, 2022 at 10:08

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.