Answer
In deep water when the value of depth $d$ is very large, the approximation $~~~v = \sqrt{\frac{gL}{2\pi}}~~~$ is appropriate.
Work Step by Step
Consider $~~tanh(\frac{2\pi d}{L})$:
$tanh(\frac{2\pi d}{L}) = \frac{sinh(\frac{2\pi d}{L})}{cosh(\frac{2\pi d}{L})}$
$tanh(\frac{2\pi d}{L}) = \frac{\frac{e^{\frac{2\pi d}{L}}-e^{{\frac{2\pi d}{L}}}}{2}}{\frac{e^{\frac{2\pi d}{L}}+e^{{\frac{2\pi d}{L}}}}{2}}$
$tanh(\frac{2\pi d}{L}) = \frac{e^{\frac{2\pi d}{L}}-e^{-{\frac{2\pi d}{L}}}}{e^{\frac{2\pi d}{L}}+e^{{-\frac{2\pi d}{L}}}}$
When $d$ is very large, then $e^{{-\frac{2\pi d}{L}}} \approx 0$
Then:
$tanh(\frac{2\pi d}{L}) \approx \frac{e^{\frac{2\pi d}{L}}-0}{e^{\frac{2\pi d}{L}}+0} = 1$
Thus, when the depth $d$ is very large:
$v = \sqrt{\frac{gL}{2\pi}~tanh(\frac{2\pi d}{L})} \approx \sqrt{\frac{gL}{2\pi}~(1)}$
Therefore, in deep water when the value of depth $d$ is very large, the approximation $~~~v = \sqrt{\frac{gL}{2\pi}}~~~$ is appropriate.
