This asymptotic representation of $\ln(1+1/x)$? I found this asymptotic approximation in an explanation for Bertrand's test, and I'm afraid I don't get it. Can someone explain what is meant by:
Using little o notation as an asymptotic approximation when $x$ tends toward infinity:
$$\ln\left(\frac{1+x}{x}\right)=\frac{1}{x}+o\left(\frac{1}{x}\right)$$
$$\left(1+\frac{1}{x}\right)^p=1+\frac{p}{x}+o\left(\frac{1}{x}\right)$$
I think I understand that in the second one, all the terms are $c/x^n$, where $n\ge 2$, so are dominated by $1/x$ but I don't understand the first one.

For the first one:
$$\ln\left(\frac{1+x}{x}\right) = \ln\left(1+\frac{1}{x}\right).$$  As $x \to\infty$, we can consider the transformation $x\to \frac{1}{y}$ and consider the Taylor expansion as $y\to 0$.  We get $$\ln\left(\frac{1+x}{x}\right) = \ln(1+y) \sim y + O(y^2),\quad\text{as }y\to 0.$$  Going back to $x$, we get $$\ln\left(\frac{1+x}{x}\right) \sim \frac{1}{x} + O\left(\frac{1}{x^2}\right)\sim\frac{1}{x} + o\left(\frac{1}{x}\right)\quad\text{as }x\to \infty.$$  Note that the change from $O$ to $o$ comes from the fact that $$f\in O\left(\frac{1}{x^2}\right)\implies \lim_{x\to\infty}\frac{f(x)}{1/x^2}<\infty \implies\lim_{x\to\infty}\frac{f(x)}{1/x} = 0\implies f\in o\left(\frac{1}{x}\right).$$
