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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:

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

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

1 Answer 1

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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.

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