Do you think there's a better way to do this?
Given that $E(x)=e^{x-1}$, $n$ iterates of $E(x)$ are defined as $E(\underbrace{E(E(...(x))...)}_{n-1\text{ times}}=E^{\circ n}(x)$.
Here's my attempt to find (or, rather, justify) $\lim\limits_{n\to\infty} E^{\circ n}(x)$.
(1) $\lvert E'(x)\lvert=\lvert e^{x-1}\lvert < 1$ if $x<1$, thus any fixed point in this range will be attractive, and so $\lim\limits_{n\to\infty} E^{\circ n}(x)=1$ if $x\le 1$.
(2) $\lvert E'(x)\lvert=\lvert e^{x-1}\lvert > 1$ if $x>1$, thus any fixed point in this range will be repulsive, and so $\lim\limits_{n\to\infty} E^{\circ n}(x)\to \infty$ if $x> 1$.
I think that something is not right with my argument because for $x=1$, $\lvert E'(x) \lvert = 1$, and so we don't know whether the point is attractive or repulsive. And for $\lvert E'(x) \lvert < 1$, there in fact are not any fixed points in the range, since the only fixed point of $E(x)$ is 1.