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Let $n \in \mathbf N^+$ and $a \in \mathbf N_{\ge 2}$. Every prime $\le 2n+1$ that doesn't divide $a$, is a divisor of $a^{(2n)!} - 1$ (by Fermat's little theorem). So we have $$\frac{\varphi(a^{(2n)!} - 1)}{a^{(2n)!} - 1} = \prod_{p \,\mid\, a^{(2n)!} - 1} \left(1 - \frac{1}{p}\right) \le \prod_{a < p \le 2n+1} \left(1 - \frac{1}{p}\right)\! \stackrel{n \to \infty}{\longrightarrow} 0,$$ where $p$ is always a prime and for the first equality we have used Euler's product formula. In particular, this shows that the limit inferior in the OP is $0$.