When $\ln(1+y) = y + o(y)$?

Question

I was reading a proof which utilize the fact that: $\ln(1+y) = y + o(y)$ https://math.stackexchange.com/a/842557/160028

I'm not so sure what is the meaning of $\ln(1+y) = y + o(y)$.

When is it true?

shouldn't it be asymptotically true (only?) when $y\to0$? Because $$\lim\limits_{y\to 0}\ln(1 +y) = \ln(1) = 0 = \lim\limits_{y\to 0}y$$

Answer 1

In the general case we have

$$f(x)=g(x)+o(g(x)),\quad \text{when}\; x\to a\in\overline{\Bbb R}\iff f(x)\sim_a g(x)$$ which's equivalent to (the case $a=\infty$ isn't very different) $$\forall \epsilon>0,\;\exists \delta>0,\;|x-a|<\delta\implies |f(x)-g(x)| <\epsilon|g(x)| $$

and in the given case we have simply $$\ln(1+y)=y+o(y)\iff\lim_{y\to0}\frac{\ln(1+y)}{y}=1$$

Answer 2

It mean that around $0$, $\ln(1+y)$ and $y$ behave in the same way. And yes, $\lim_{y\to 0}\ln(1+y)=\lim_{y\to 0}y$. For exemple, to compute $\lim_{y\to 0}\frac{\ln(1+y)}{y}$ you know that $\ln(1+y)=y+o(y)$ (we note also $\ln(1+y)\sim y$ if $y\to 0$) and so,

$$\lim_{y\to 0}\frac{\ln(1+y)}{y}=\lim_{y\to 0}\frac{y}{y}=1.$$