= Solution
Integrate $e^{iz}/z$ around the upper half-disc with the upper half-disc of radius $\varepsilon$ about <zero> removed. The real segments run from $-R$ to $-\varepsilon$ and $\varepsilon$ to $R$; the small semicircle is clockwise. There are no poles inside this indented <contour>.
The large arc contribution tends to <zero>: its modulus is at most $\int_0^\pi e^{-R\sin\theta}\,d\theta$, which tends to <zero> by <dominated convergence>. On the small arc $z=\varepsilon e^{i\theta}$, $\theta$ decreases from $\pi$ to <zero>, so its integral tends to $\int_\pi^0i\,d\theta=-i\pi$. The <contour> theorem consequently gives
$$
\operatorname{PV}\int_{-\infty}^{\infty}\frac{e^{ix}}x\,dx=i\pi.
$$
Taking imaginary parts is legitimate for each truncated integral. The imaginary integrand $\sin x/x$ is continuous at <zero> after assigning its limiting value one, so the inner indentation can be removed. Therefore
$$
\boxed{\lim_{R\to\infty}\int_{-R}^{R}\frac{\sin x}{x}\,dx=\pi.}
$$
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