$f(0)=0 \neq f'(0) \Rightarrow \frac{\delta(x)}{f(x)} = \frac{\delta(x)}{x f'(0)}$? In this reference (Eq. 15 and sentence below) the following property of Dirac's delta is used:
$$ k_n \frac{ \delta(x)}{\sin(k_n x)} \rightarrow k_n \frac{ \delta(x)}{x \, k_n} = \frac{ \delta(x)}{x} $$
How can one justify this relation?
This obscure step is necessary because the result on the right side must be independent of the constant $k_n$.
I know that $g(x) \delta(x) = g(0) \delta(x)$ for valid test functions $g$, but in this case we encounter $g(x) = \frac{1}{f(x)}$ with $f(x)=0$. I'm not familiar with this case.
For consistency, I'm using the standard definition as a distribution of the delta. I imagine in this particular case an extended definition would be needed in order to include this type of undefined test functions at the origin, in a way that is compatible with the standard one. Moreover, the distribution $D_{\frac{\delta(x)}{f(x)}}[h]$ has to be well-defined for test functions $h(0) = 0 \neq h'(0)$, because later in the manuscript (Eq. 20) the following is used: $\sin(k_n x) \frac{\delta(x)}{x} = k_n \delta(x)$
I believe there is already a mainstream way of handling this type of issues, but I can't find it. If it helps, my guess is that $\frac{\delta(x)}{f(x)} = \frac{\delta(x)}{x f'(0)}$ and $\frac{h(x) \delta(x)}{f(x)} = \frac{h'(0) \delta(x)}{f'(0)}$, where I used the lowest-order term in the Taylor expansion at the origin.
 A: In the sense of distributions, such a quantity is only defined on the dual of a subspace of a function space where $V = \{\varphi\in C_0^\infty(\Bbb{R})| \varphi(0)=0\}$. However, it does make sense to talk about the following equality in a more general function space $V$ - does there exist a distribution $g\in V^*$ such that for $u\in V$
$$u(x)g(x) = k\delta(x)$$
when $u(0) = 0 \neq u'(0)$ ? Well yes, the derivative of $\delta$ does the job perfectly
$$u(x)\delta'(x) \equiv - u'(0) \delta(x) = k\delta(x)$$
And so the question becomes, is the action of $\frac{\sin(k_n x)}{k_n}$ on $\delta'$ the same as the action of $x$ on $\delta'$ ? Absolutely because they both share values of their $0$th and $1$st derivatives:
$$\frac{\sin(k_n x)}{k_n}\delta'(x) \equiv -\delta(x) \equiv x\delta'(x)$$
In other words, taking $g = -\delta'$, we get the equality of distributions we are after
$$\frac{k_n \delta(x)}{\sin(k_n x)} = \frac{\delta(x)}{x}$$
however this is only defined on the subspace where $\varphi(0)=0$. This can always be regularized with the homogenized version of the function on the larger function space
$$\left\langle\frac{\delta(x)}{x},\varphi\right\rangle \equiv \left\langle \delta, \frac{\varphi(x) - \varphi(0)}{x}\right\rangle$$
which is not a proper linear functional technically. Further regularization with higher order derivatives vanishing would use higher order derivatives of the delta as intermediaries like we did here with the first derivative.
