Please let me know if this proof is OK.
Problem statement: Given that $E_\mu(x) = \mu e^x$, where $0 < \mu < 1/e$, show that if $q_\mu < p_\mu$ are fixed points, where $q_\mu$ is attractive and $p_\mu$ is repelling, then $\lim\limits_{\mu\to 0^+} q_\mu = 0$ and $\lim\limits_{\mu\to 0^+} p_\mu=\infty$.
My attempt:
Since we know that $p_\mu > q_\mu$ is repelling and $q_\mu$ is attractive, we know that $E'(p_\mu) > 1$ and $E'(q_\mu) < 1$. For this range of values of $\mu$, we can express: $E_\mu(x) = e^{x-\varepsilon}$, for any $\varepsilon > 1$. Now, $E'(\overline{x}) = E(\overline{x})$, and $E(q_\mu) = E'(q_\mu) = e^{q_\mu}e^{-\varepsilon} = q_\mu$. As $\mu \to 0^{+}$, $\varepsilon \to \infty$. Thus we have $\lim\limits_{\mu\to 0^+} q_\mu = e^{q_\mu} e^{-\varepsilon} < 1$, which implies that $q_\mu \to 0.\\ \\$ Similarly, $\lim\limits_{\mu\to 0^+} p_\mu = e^{p_\mu} e^{-\varepsilon} > 1$ implies that $p_\mu$ must approach $\infty$ faster than $\varepsilon$.
I'm a bit concerned about the final part, where I make conclusions about the limits.