By the definition, we know that,
\(\tanh \left( {\frac{{2\pi d}}{L}} \right) = \frac{{{e^{\left( {\frac{{2\pi d}}{L}} \right)}} - {e^{ - \left( {\frac{{2\pi d}}{L}} \right)}}}}{{{e^{\left( {\frac{{2\pi d}}{L}} \right)}} + {e^{ - \left( {\frac{{2\pi d}}{L}} \right)}}}}\)
As we need to find what happens to the velocity as the depth increases
Take the limit on both sides of the above function.
\(\mathop {\lim }\limits_{d \to \infty } \tanh \left( {\frac{{2\pi d}}{L}} \right) = \mathop {\lim }\limits_{d \to \infty } \frac{{{e^{\left( {\frac{{2\pi d}}{L}} \right)}} - {e^{ - \left( {\frac{{2\pi d}}{L}} \right)}}}}{{{e^{\left( {\frac{{2\pi d}}{L}} \right)}} + {e^{ - \left( {\frac{{2\pi d}}{L}} \right)}}}}\)
As in the above function, the situation is \(\frac{\infty }{\infty }\) . Apply L’hopital’s rule:
\(\begin{aligned}\mathop {\lim }\limits_{d \to \infty } \frac{{{e^{\left( {\frac{{2\pi d}}{L}} \right)}} - {e^{ - \left( {\frac{{2\pi d}}{L}} \right)}}}}{{{e^{\left( {\frac{{2\pi d}}{L}} \right)}} + {e^{ - \left( {\frac{{2\pi d}}{L}} \right)}}}} & = \mathop {\lim }\limits_{d \to \infty } \frac{{4\pi {e^{4\pi d}}}}{{4\pi {e^{4\pi d}}}}\\ & = \mathop {\lim }\limits_{d \to \infty } 1\\ & = 1\end{aligned}\)
Therefore, we get,
\(\mathop {\lim }\limits_{d \to \infty } \tanh \left( {\frac{{2\pi d}}{L}} \right) = 1\)
Now solve the given velocity as shown below:
\(\begin{aligned}v & = \sqrt {\left( {\frac{{gL}}{{2\pi }}} \right)\tanh \left( {\frac{{2\pi d}}{L}} \right)} \\ & = \sqrt {\left( {\frac{{gL}}{{2\pi }}} \right) \cdot 1} \\ \approx \sqrt {\left( {\frac{{gL}}{{2\pi }}} \right)} \end{aligned}\)
Thus, as the depth \(d\) increases the function \(\tanh \left( {\frac{{2\pi d}}{L}} \right) \approx 1\) , therefore \(v \approx \sqrt {\left( {\frac{{gL}}{{2\pi }}} \right)} \).
