= Solution
Let $\phi$ be a <test function> supported in $[-R,R]$. Symmetric cancellation gives
$$
\operatorname{PV}\frac1x(\phi)=\int_0^R\frac{\phi(x)-\phi(-x)}x\,dx.
$$
The integrand is bounded by $2\|\phi'\|_\infty$, so the limit exists and
$$
\left|\operatorname{PV}\frac1x(\phi)\right|\leq2R\|\phi'\|_\infty.
$$
This is a continuous linear functional on every fixed-support test space, hence a <distribution>. It also specifies its local order as at most one.
For the boundary value, split the ordinary complex denominator:
$$
\frac1{x+i\varepsilon}=\frac{x}{x^2+\varepsilon^2}-i\frac{\varepsilon}{x^2+\varepsilon^2},\qquad\varepsilon>0.
$$
The real part paired with $\phi$ is
$$
\int_0^R\frac{x[\phi(x)-\phi(-x)]}{x^2+\varepsilon^2}\,dx.
$$
It converges to the principal value by <dominated convergence theorem>, using the same bound $2\|\phi'\|_\infty$. In the second part set $x=\varepsilon t$:
$$
\int_\mathbb R\frac{\varepsilon\phi(x)}{x^2+\varepsilon^2}\,dx=\int_\mathbb R\frac{\phi(\varepsilon t)}{1+t^2}\,dt\longrightarrow\pi\phi(0).
$$
Here $\|\phi\|_\infty/(1+t^2)$ is an integrable majorant. The correct <Sokhotski–Plemelj formula> is therefore
$$
\boxed{\frac1{x+i0}=\operatorname{PV}\frac1x-i\pi\delta.}
$$
The resulting functional is continuous, with bound $2R\|\phi'\|_\infty+\pi\|\phi\|_\infty$, so it too is a <distribution>.
\b[The printed PDF has the wrong sign on its delta term.] For a real even nonnegative <test function> with $\phi(0)>0$, the principal value is zero while the imaginary part of every regularized integral is negative. Its limit is $-\pi\phi(0)$, disproving the printed plus sign directly. A plus sign instead belongs to the boundary value $(x-i0)^{-1}$.
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