Limiting cases for $a/(b+a)$

Question

Maybe a simple question but I am a bit stuck with it. $\frac{a}{b+a}$ is a part of the equation that I have to solve in two limiting cases.

For $b\ll a$ we can skip $b$ and it ends up in $\frac{a}{a}$. (That is, $\frac{a}{a+b}\to \frac{a}{a}$ as $b\to0$).

But what in case when $b\gg a$? Should the asymptotic approximation to $\frac{a}{a+b}$ be $\frac{a}{b}$ or $\frac{1}{b}$?

Answer 1

Here if $b>>a$ then in the limiting case it tends to zero. So any form $\frac ab$ or $\frac 1b$ can be taken as per your choice. But I think $\frac ab$ is better choice.

Answer 2

I think you need to consider the case where $b\to\infty$, or if you prefer $b\gg a \implies \frac{b}{a}\gg 1$, or $\frac{b}{a}\to\infty \implies \frac{a}{b}\to 0$ (you can think it as $\frac{a}{b}\ll 1$), such that $\frac{a}{a+b}\to 0$. It seems more consistent to me.