I've been trying to show that
$$\lim_{t\to \infty} \dfrac{e^{ixt}}{x-i0}=2\pi i \delta(x).$$
For that I've used the fact that
$$\dfrac{1}{x-i0}=\lim_{\epsilon\to 0^+} \dfrac{1}{x-i\epsilon}=i\pi \delta(x)+\operatorname{Pv}\dfrac{1}{x},$$
and the fact that if $f$ is a $C^\infty$ function, $\xi$ a distribution and $\phi$ a test-function, by definition we have $(f\xi,\phi)=(\xi,f\phi)$. In that setting we have that the distribution inside the limit is defined by
$$\left(\dfrac{e^{ixt}}{x-i0},\phi(x)\right)=\left(\dfrac{1}{x-i0},e^{ixt}\phi(x)\right).$$
Hence applying the formula I've mentioned, we have
$$\left(\dfrac{e^{ixt}}{x-i0},\phi(x)\right)=i\pi \phi(0)+\lim_{\eta\to 0}\left(\int_{-\infty}^{-\eta}+\int_{\eta}^{\infty}\right)\dfrac{e^{ixt}\phi(x)}{x}dx,$$
the point is that using contour integration we can easily see that that limit of the integrals is just $i\pi\phi(0)$ provided that $t > 0$. In that sense, the action of the distribution doesn't depend on $t$ at all, it is just
$$\left(\dfrac{e^{ixt}}{x-i0},\phi(x)\right)=2\pi i(\delta(x),\phi(x)),$$
and hence taking the limit in the pointwise fashion since the $t$ dependence vanishes we get the initial result.
What I'm wondering here is: is my conclusion correct? The distribution $e^{ixt}/x-i0$ doesn't really depend on $t$ at all? Again, my only supposition was that $t > 0$, I never had to take the limit in order to get the result.