Is there a sense in which one could expand $\frac{\sin (x/\epsilon )}{x} $ in powers of $\epsilon $? A standard representation of the $\delta $-distribution is
$$
\pi \delta (x) = \lim_{\epsilon \searrow 0} \frac{\sin (x/\epsilon )}{x}
$$
Is there a sense in which this could be seen as the leading term in an expansion of $\sin (x/\epsilon ) /x$ in positive powers of $\epsilon $, presumably with distributions as coefficients? If so, is it possible to give the expansion explicitly?
 A: If you are interested in Laurent series, the expansion is
$$\frac{\sin (x/\epsilon)}x=\frac{1}{\epsilon }-\frac{x^2}{6 \epsilon ^3}+\frac{x^4}{120 \epsilon ^5}-\frac{x^6}{5040 \epsilon ^7}+\frac{x^8}{362880 \epsilon ^9}+O\left(\frac{1}{\epsilon^{11} }\right)=\sum_{n=0}^\infty \frac {(-1)^{2n}x^{2n}}{(2n+1)!\epsilon^{2n+1}}$$
A: The expansion around point $a$ in positive powers exists, but what is the general form of the term and whether it is convergent requires further research.
$$\frac{\sin (x/\epsilon)}x=\frac{\sin \left(\frac{x}{a}\right)}{x}-\frac{(\epsilon -a) \cos \left(\frac{x}{a}\right)}{a^2}+\frac{(\epsilon -a)^2 \left(2 a \cos \left(\frac{x}{a}\right)-x \sin \left(\frac{x}{a}\right)\right)}{2 a^4}$$
$$+\frac{(\epsilon -a)^3 \left(-6 a^2 \cos \left(\frac{x}{a}\right)+x^2 \cos \left(\frac{x}{a}\right)+6 a x \sin \left(\frac{x}{a}\right)\right)}{6 a^6}$$
$$+\frac{(\epsilon -a)^4 \left(24 a^3 \cos \left(\frac{x}{a}\right)-36 a^2 x \sin \left(\frac{x}{a}\right)+x^3 \sin \left(\frac{x}{a}\right)-12 a x^2 \cos \left(\frac{x}{a}\right)\right)}{24 a^8}$$
$$+\frac{(\epsilon -a)^5 \left(-120 a^4 \cos \left(\frac{x}{a}\right)+240 a^3 x \sin \left(\frac{x}{a}\right)+120 a^2 x^2 \cos \left(\frac{x}{a}\right)-x^4 \cos \left(\frac{x}{a}\right)-20 a x^3 \sin \left(\frac{x}{a}\right)\right)}{120 a^{10}}$$
$$+O\left((\epsilon -a)^6\right)$$
