For a sequence $(x\_{\alpha})$ of surreal numbers indexed by the set of all ordinal numbers, we say that $\lim x\_{\alpha}=l$ ($l$ is a surreal number) if for each surreal $\epsilon>0$, there exists an ordinal $\beta$ such that $|x\_{\alpha}-l|<\epsilon$ for each ordinal $\alpha>\beta$.


Consider the sequence $x\_{\alpha}=(1+1/\alpha)^\alpha$, where for each surreal $x,y$, $x^y$ is defined by $\exp(y\log x)$.


Q1: What is the normal form of $x\_{\omega}$? Is its real part equals to $e$?


Q2: What is $\lim x\_{\alpha}$?



The surreal version of $e$
$\DeclareMathOperator{\ee}{e}$If $\varepsilon$ is an infinitesimal surreal, the quantity $\log(1+\varepsilon)$ is actually equal to the formal sum *à la Hahn series* $\sum \limits\_{n \in \mathbb{N}} \frac{(-1)^n\varepsilon^{n+1}}{n+1}$, and $\log(1+\varepsilon) - \varepsilon$ is negligeable with respect to $\varepsilon$. So the proof for real numbers can be applied here to show that the sequence converges to $\ee$. 


As for $x\_{\omega}$, this is $\exp(1 - \frac{1}{2\omega} + \frac{1}{3\omega^2} - ...)$ where $\exp(- \frac{1}{2\omega} + \frac{1}{3\omega^2} - ...)$ is infinitesimally close to $1$ because $a:= -\frac{1}{2\omega} + \frac{1}{3\omega^2} - ...$ is an infinitesimal. So the real part of $x\_{\omega}$ is $\ee$.




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For now I don't see what its normal form because some combinatorial cleverness seems to be required in unfolding $\sum \limits\_{n \in \mathbb{N}} \frac{a^n}{n!}$. I'll edit this answer if I find something. In the meantime you can try to find it too: we know that exponents of $\omega$ in the $a^n$ and $\exp(a)$ will be negative integers. It might be easier to compute the exponential sum directly if you can find relations between the coeffifients $q\_{n,k}$ of $\omega^{-k}$ in $a^n$ for different values of $n$. 


