For a given $x > 0$, let $a_n$ be the sequence defined by $a_1=x$ for $n = 1$ and $a_n = x^{a_{n−1}}$ for $n\ge 2$. Find the largest $x$ for which the limit $\displaystyle\lim_{n\to\infty} a_n$ converges.
In order for $\displaystyle\lim_{n\to\infty}a_n$ to have a limit $L$, it must hold that $x^L=L$ or $x=L^{1/L}$. Therefore, the problem is equivalent to maximize $f(L)=L^{1/L}$.
$$\frac{df}{dL}=\frac{d}{dL}e^{\frac{\ln L}{L}}=L^{1/L}\left(\frac{1}{L^2} - \frac{\ln L}{L^2}\right)$$
Setting this be $0$ leads to $L=e$ and accordingly $x=\boxed{e^{1/e}}$
We also need to make sure the value is indeed a maximum. For this, let's check its second derivative at $L=e$.
$$\frac{d^2f}{dL^2}|_e=L^{\frac{1}{L}-4}\left(-3L+\ln^2L+2(L-1)\ln{L}+1\right)|_e=-e^{1/e-3}<0$$