Open-loop TF has no poles at zero implies there is nonzero offset $\newcommand{\Gol}{G_{\rm ol}}\newcommand{\Gcl}{G_{\rm cl}}\newcommand{\llim}{\lim_{s \overset{\mathbb{R}}{\to} 0^+}}$I believe the following is true, but I have not been able to prove it:

Consider a closed-loop system with open-loop transfer function $\Gol$ and closed-loop transfer function $\Gcl$, both stable. If $\lim_{\omega\to0^+}|G(j\omega)| \neq \infty$ then the offset upon a step change of the set point is not zero. For simplicity, assume that the transfer function of the sensor is $G_m(s) = 1$.

Attempt to prove this. The open-loop transfer function is $\Gol(s) = G_c(s)G_a(s)G_p(s)$, where $G_c$, $G_a$ and $G_p$ are the TFs of the controller, actuator and controlled plant respectively. Then, the closed-loop TF is $\Gcl(s) = \frac{\Gol(s)}{1 + \Gol(s)}$. Since $\Gcl$ is assumed to be stable, we can apply the final value theorem on $\Gcl(s)/s$. This means that we can determine the offset as follows
$$
{\rm Offset} := \lim_{t\to\infty} 1 - x^{\rm step}(t) = 1 - \llim s \frac{\Gcl(s)}{s} = 1 - \llim\Gcl(s).
$$
The notation "$\llim$" refers to the limit over $s\in\mathbb{R}$ as $s\to0^+$. In order to have zero offset, the following must hold:
$$
\llim \Gcl(s) = 1 
\Leftrightarrow
\llim \frac{\Gol(s)}{1 + \Gol(s)} = 1.
$$
Firstly, am I right in assuming that this can only happen if
$$
\llim|\Gol(s)| = \infty?
$$
My question. If the limit of $\Gol$ as $s{}\to{}0$ (for any sequence of complex numbers) exists, then $\llim|\Gol(s)|$ is equal to $\lim_{\omega\to 0^+}|\Gol(j\omega)|$ (for $\omega\in\mathbb{R}$), but in general it may not be true that the above limit exists, right? What are some reasonable assumptions under which we can establish a correspondence between $\llim|\Gol(s)|$ and $\lim_{\omega\to 0^+}|\Gol(j\omega)|$, that is, the limit over the reals and limit over the imaginary numbers?
My motivation to prove the above claim comes from Bode plots. I wonder whether it is possible to tell whether a closed-loop system has offset by looking at the Bode plot of its open-loop transfer function. 
 A: To answer your first question, the following are equivalent

*

*$\lim_{\nu}\frac{a_\nu}{1 + a_\nu} = 1$

*$\lim_{\nu} a_\nu = \pm\infty$
Proof. It follows from the fact that $\frac{a_\nu}{1 + a_\nu} = \frac{1}{1+a_\nu^{-1}}$.
For the second part, we can assume without great loss of generality that
\begin{align}
G_{\mathrm{ol}}(s) = \frac{A(s)}{B(s)},
\end{align}
where $A, B:D\to\mathbb{C}$, $0$ is either inside $D$ or it is at its boundary and $A$ is locally bounded in a neighbourhood of $0$ relative to $D$. Then we have
\begin{align}
\lim_{s \overset{\mathbb{R}}{\to} 0^+} \left|G_{\mathrm{ol}}(s)\right| = \infty
\Leftrightarrow
\lim_{s \overset{\mathbb{R}}{\to} 0^+} \left|\frac{A(s)}{B(s)}\right| = \infty.
\tag{1}\label{1}
\end{align}
Under the boundedness assumption on $A$, Equation \eqref{1} holds iff
\begin{align}
\lim_{s \overset{\mathbb{R}}{\to} 0^+} \left|B(s)\right| = 0.\tag{2}\label{2}
\end{align}
The assumption that $G_{\mathrm{ol}}(s) = A(s)/B(s)$, where $A$ and $B$ are not necessarily polynomials, encompasses a very large class of transfer functions, including ones with exponentials in the denominator.
If we assume that the limit
$$\lim_{s \overset{\mathbb{C}}{\to} 0^+} \left|B(s)\right|$$
exists, or that $B$ is continuous at zero, then Equation \eqref{2} holds iff
\begin{align}
\lim_{\omega\to0^+}|B(j\omega)|=0,
\end{align}
in which case, in light of the local boundedness of $A$,
\begin{align}
\lim_{\omega\to0^+}|G_{\mathrm{ol}}(j\omega)|=\infty,
\end{align}
which is something that can be readily checked given the Bode plot of $G_{\mathrm{ol}}$.
