Limit approximation for $\pi$ in the four fours puzzle? The four fours puzzle is a recreational math puzzle whose aim is to express whole numbers using four occurrences of the digit 4 and a specified set of operators. A common variety permits the following: 


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*Constants: decimals (e.g. 44, which uses two 4s), leading decimal point (e.g. $.4 = \frac{2}{5}$) and recurring decimals (e.g. $.\overline{4} = \frac{4}{9}$).

*Unary operators: negation $-x$, square root $\sqrt{x}$ and integer factorial $x!$.

*Binary operators: +, -, ×, ÷, exponentiation $x^y$ and arbitrary roots $\sqrt[y]{x}$.


If we extend the puzzle to arbitrary reals then it is clear by countability that some numbers can't be expressed precisely (though by Gelfond–Schneider some transcendental numbers can). However, for some potentially unexpressable numbers it is possible to express arbitrarily close estimates. For example 
$$e_n = \sqrt{\left(\frac{4!^{(n)} + \sqrt{4}}{4!^{(n)}} \right) ^{4!^{(n)}}}$$
where
$$4!^{(n)} = 4\overbrace{!...!}^{n\textrm{ times}}$$ 
is a sequence of four fours expressions whose limit is $e$.
Is there a similar expressible limit expression for $\pi$? There are a number of relevant limits at Pi: Limit representations but I can't think of how to express any of them using just four 4s.
Note that proving the hypothesis in Repeated Factorials and Repeated Square Rooting would show that arbitrarily close estimates are possible for every number using just one 4, but without necessarily providing them explicitly.
Note also that if we permit usage of the $\ln$ operator, then we can approximate any number using just three 4s. For example, the following is a sequence of four fours expressions whose limit is $a \geq 1$:
$$a_n = \overbrace{\sqrt{\sqrt{\cdots \sqrt{- \frac{\ln ( \ln \overbrace{\sqrt{\sqrt{\cdots \sqrt{4}}}}^{\lfloor a^{2^n}\rfloor \textrm{ times}} / \ln 4 )}{\ln \sqrt{4}}}}}}^{n \textrm{ times}}$$

If you extend the domain of the factorial operator to the reals, then:
$$4\cdot \left(\frac{\sqrt{\sqrt{\sqrt{\ldots\sqrt{\sqrt{4}}}}}}{\sqrt{4}}!\right)^{\sqrt{4}}$$
approaches $\pi$ as the number of square roots increases.
