= Solution
The <Beth numbers> form a strictly increasing continuous sequence: strictness comes from <Cantor theorem>, and at a limit $\delta$ one has $\beth_\delta=\sup_{\xi<\delta}\beth_\xi$. A cofinal sequence in $\delta$ of length $\operatorname{cf}(\delta)$ therefore induces a cofinal sequence in $\beth_\delta$, giving $\operatorname{cf}(\beth_\delta)\le\operatorname{cf}(\delta)$.
Conversely, let $\{a_i:i<\operatorname{cf}(\beth_\delta)\}$ be cofinal in $\beth_\delta$. For each $i$ choose $\xi_i<\delta$ with $a_i<\beth_{\xi_i}$. The indices must be cofinal in $\delta$; otherwise all $a_i$ would be bounded by a single smaller Beth number. Thus $\operatorname{cf}(\delta)\le\operatorname{cf}(\beth_\delta)$, and
$$
\boxed{\operatorname{cf}(\beth_\delta)=\operatorname{cf}(\delta).}
$$
This is the <cofinality of a continuous cardinal hierarchy> at a nonzero limit index; the limit hypothesis is essential to the supremum argument.
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