Why do we use big Oh in taylor series?
The notion of Big O, here, is to give an approximation/upper bound in the neighborhood of the value. It means that if you have a numerical approximation in a small neighborhood of x then higher order terms rapidly go to zero with small perturbations. If I move $.001$ from $x$ then $.001$ to the seventh is next largest term in the series and is, indeed, very small.
The second two lines of the formal definition from wikipedia are what you want to think about in terms of a taylor series approximating a periodic function.
The Big Oh notation here is not the same as it is used, e.g., in the study of algorithms in computer science. It is actually about the behaviour of $f$ close to a point (zero in this case). We could define $f(x) = O\bigl(g(x)\bigr)$ if there exists $\varepsilon > 0$ such that for $|x| < \varepsilon$ we have $f(x) \leq c g(x)$ for some constant $c$.
As for why we use this notation: Sometimes, higher-order terms are not really relevant for describing the local behavior of functions. E.g., from $$\sin x = x- \frac{x^3}{6} + \frac{x^5}{120} + O(x^7)$$ it follows that
- $\sin (0) = 0$,
- $\sin' (0) = 1$,
- $\sin'' (0) = 0,$
- $\sin^{(3)} (0) = -1,$
- $\sin^{(4)} (0) = 0,$
- $\sin^{(5)} (0) = 1,$
- $\sin^{(6)}(0) = 0$.
Note, however, that it does not imply that $\sin^{(7)} (0)= -1$.
Another reason for using the Big Oh notation is that it basically encompasses Taylor's Theorem, but in a potentially more readable way: Whenever $f$ is $n$-times continuously differentiable at zero, we have $$f(x) = f(0) + f'(0) \cdot x + \frac{f''(0)}{2} \cdot x^2 + \cdots + \frac{f^{(n)}(0)}{n!} \cdot x^n + O(x^{{n+1}}).$$
We can write where the big Oh takes place to be clearer: $$ \sin x = x + \underset{x \to 0}{\mathcal O} \left( x^3 \right)$$ or $$ \sqrt{1+ \frac{1}{n}} = 1 + \frac{1}{2n} + \underset{n \to +\infty}{\mathcal O} \left( \frac1{n^2} \right)$$