While reading some old notes on contour integration, I noticed the author uses series expansion:
$$\frac{\sinh sx}{\sinh sa}=\frac{sx(1+\frac{1}{6}s^2x^2+\cdots)}{sa(1+\frac{1}{6}s^2a^2+\cdots)}=\frac{x}{a}(1+\frac{1}{6}s^2(x^2-a^2)+\cdots)$$ for small $s$
This may be a silly question, but could someone tell me the justification behind this?
The first part is an application of the Taylor series of $\sinh(z)$, which is
$$\sinh(z)=z+\frac{z^3}{3!}+\frac{z^5}{5!}+\frac{z^7}{7!}...$$
Collecting $z$ we get
$$\sinh(z)=z(1+\frac{z^2}{3!}+\frac{z^4}{5!}+\frac{z^6}{7!}...)$$
which corresponds to the first expression provided in the question.
The second expression, where the ratio of the two series $1+\frac{x^2}{3!}+\frac{x^4}{5!}+\frac{x^6}{7!}...$ and $1+\frac{a^2}{3!}+\frac{a^4}{5!}+\frac{a^6}{7!}...$ is substituted by $1$ plus their difference, is explained by a rough approximation. This is based on the fact that, if we have two numbers $1+p$ and $1+q$ with $p<<1$ and $q<<1$, their ratio $1+r$ can be estimated as
$$\frac{(1+p)}{(1+q)}=(1+r)$$
$$(1+p)=(1+r)(1+q) \\ =1+q+r+qr$$
Neglecting the "small" term $qr$, we get $1+p=1+q+r$, and then
$$1+r=1+p-q$$
In the equation shown in the OP, this approximation can be applied if $s$ is sufficiently small.