Suppose we have a Taylor expansion of $f(x) = \frac{1}{1+x}$ at x = 0 of degree 0, so with Lagrange remainder:
$$f(x) = 1 + xf'(\xi)$$ for some $\xi \in (0,x)$ (suppose $x$ is positive)
Now, in class this was written as:
$$f(x) = 1 + O(x)$$
What I don't understand is why can we treat $f'(\xi)$ as a constant? Isn't it possible that $f'(x)$ grows very fast and for a large enough $x$, $f'(\xi)$ will "overpower" $x$?
thanks.
On
$f=O(g)$ means that "close enough" to $a$(this case, $0$) the absolute value of $f$ is bounded from above by the absolute value of $g$ times constant.
This here we get that "close enough" to $0$ we have $|xf'(x)|$ is bounded by some constant times $|x|$, this is only near $0$. In this case we can compute to see that any closed interval from $-a$ to $a$ is close enough when $|a|<1$.
You are not considering very large $x$, you are considering the asymptotics for $x\to 0$. In that sense you need to find a bound $M$ for $f'(ξ)$ on some interval $[-r,r]$ around $x=0$.
Then on that interval you get $|f(x)-1|\le M|x|$ for $|x|\le r$, which is the definition for $f(x)=1+O(x)$.
Here one can compute directly $$ f(x)-1=-\frac{x}{1+x}\implies |f(x)-1|\le\frac1{1-r}|x| ~\text{ for }~ |x|\le r. $$