Taylor expansion of ln(1+x) and small O http://people.math.sc.edu/girardi/m142/handouts/10sTaylorPolySeries.pdf
The Taylor expansion of $\ln(1+x)$ is $\sum_{n=1}^{\infty} (-1)^{n-1}\frac{x^n}{n}$. Is it true that we can think of $\ln(1+x) = x+o(x^2)$, what does this mean pricesly? I find that such thing is an usual trick throughout mathematical analysis, but I can not find any related material on Baby Rudin. Does any one have any rigorous reference on this?
 A: What you have written is not true. What is true is that $$\ln(1+x) = x + O(x^2)$$, or $$¢\ln(1+x) = x + o(x)$$ (the former implies the latter, so is a
stronger statement).
We write $$f(x) = o(g(x)) $$ iff for all $$\epsilon >0$$  there is some $$\delta>0$$ such that if $|x| < \delta $, then $$|f(x)| \le \epsilon |g(x)| $$ 
We write $$f(x) = O(g(x))$$ iff there exists some $M$   and  some $\delta>0$
 such that if $$ |x| < \delta $$, then $$|f(x)| \le M |g(x)|$$.
To see the specifics:
$$\ln(1+x) = x - {x^2 \over 2} + x^3\sum_{n=3}^{\infty} (-1)^{n-1}\frac{x^{n-3}}{n} $$. For $$|x| < {1 \over 2}$$, it is easy to show that the
function $$h(x) = \sum_{n=3}^{\infty} (-1)^{n-1}\frac{x^{n-3}}{n}$$ is bounded, say $|h(x)| \le K$.
We have $\ln(1+x) -x = - {x^2 \over 2} + x^3 h(x)$ and so, for $|x| < { 1\over 2}$, we have
$$|\ln(1+x) -x | \le x^2 ({1 \over 2}+{1 \over 2} K)$$, and so $$x \mapsto \ln(1+x) -x$$ is $$O(x^2)$$.
Given $\epsilon>0$, choose $$|x| < {1 \over {1 \over 2}+{1 \over 2} K} \min(\epsilon, {1 \over 2})$$ to get
$$|\ln(1+x) -x | \le \epsilon |x|$$, and so
$$x \mapsto \ln(1+x) -x$$ is $o(x)$.
A: 
Is it true that $\ln (1 + x) = x + o(x^2)$?

No, but it's almost true. It is true that
$$ \ln(1+x) = x + O(x^2)$$
for small $x$. This is called "big oh" notation. One says that a function $f(x)$ is in $O(x^2)$ if there is some constant $C$ and some constant $x_0$ such that
$$ f(x) \leq C x^2 $$
for $\lvert x \rvert < x_0$. For this function, this is a direct application of Taylor's Theorem.
