 Solution

WriteFor a static observer, $d\tau =\sqrt{f(r)}dt$. Conservation of the photon's Killing energy therefore gives the gravitational redshift between static Schwarzschild observersFor the stated distant observer, $r\gg r_{\mathrm{st}}$ and $f(r)\simeq 1$, soAs $r_{\mathrm{st}}\to 2M^{+}$, the observed wavelength tends to infinity: light emitted from a static source arbitrarily near the Schwarzschild event horizon is infinitely redshifted at infinity.