Using Sokhotsky's formulae prove the following limits when $t \rightarrow + \infty$
$$\frac{e^{ixt}}{x - i0} \rightarrow 2 \pi i \delta(x)$$
and
$$\frac{e^{ixt}}{x + i0} \rightarrow 0.$$
Where Sokhotsky's formulae are
$$\frac{1}{x + i0} = -i \pi \delta(x) + p.v.(\frac{1}{x}),$$
$$\frac{1}{x - i0} = i \pi \delta(x) + p.v.(\frac{1}{x}).$$
We start with expanding the expressions: $$ \frac{e^{ixt}}{x \pm i0} = e^{ixt} ( \mp i\pi\,\delta(x) + \text{pv}\frac{1}{x} ) = \mp i\pi\,e^{ixt}\delta(x) + (\cos xt)\text{pv}\frac{1}{x} + i (\sin xt)\text{pv}\frac{1}{x} . $$
For the first term we use $f(x)\,\delta(x) = f(0)\,\delta(x)$ to get $\mp i\pi\,e^{ixt}\delta(x) = \mp i\pi\,\delta(x).$
I haven't been able to prove it yet, but I am sure that the second term tends to $0$ as $t \to \pm\infty$ since the quick oscillations make things cancel, even at $x=0$.
The third term tends to $i\pi\,\delta(x)$. See for example your own question about this limit. Here the oscillations do not cancel at $x=0$ but instead give a spike of height $t$.
Thus the limits are $\mp i\pi\,\delta(x) + i\pi\,\delta(x).$