Can an ARMA(4,1) be the same as an ARMA(3,0)? I would really appreciate it if someone could explain the reasoning behind whether this can be true or not.
 A: No. And the reasoning why doesn't have anything to do with limits or approximations.
If you take an ARMA(4,1) model, the only way it could be an ARMA(3) model is if its AR polynomial had a root in common with its MA polynomial. If there was a common root, you could divide both sides of the model by it, and both $p$ and $q$ would reduce by $1$. Most of the definitions that I've seen restrict this behavior.
For example, in "Intro to Time Series and Forecasting," their definition is

$\{X_t\}$ is an ARMA(p,q) process if $\{X_t\}$ is stationary and if
  for every $t$,  $$ X_t - \phi_1 X_{t-1} - \cdots - \phi_p X_{t-p} =
 Z_t + \theta_1 Z_{t-1} + \cdots + \theta_q Z_{t-q}, $$ where $\{Z_t\}
 \sim \text{WN}(0,\sigma^2)$ and the polynomials $(1 - \phi_1 z -
 \cdots - \phi_p z^p)$ and $(1 + \theta_1z + \cdots + \theta_q z^q)$
  have no common factors.

Say, to the contrary, that your model had $z_1 = z_{1,\phi} = z_{1,\theta}$ as a common root. Then
\begin{align*}
\frac{\phi(z)}{\theta(z)}&= \frac{\phi_p(z-z_{1,\phi})(z-z_{2,\phi})\cdots(z-z_{p,\phi})}{\theta_q(z-z_{1,\theta})\cdots(z-z_{q,\theta})}\\
&= \frac{\phi_p(z-z_{2,\phi})\cdots(z-z_{p,\phi})}{\theta_q(z-z_{2,\theta})\cdots(z-z_{q,\theta})}.
\end{align*}
You could write down your larger model, divide both sides by the common factor, and the result would be a smaller model. And in your specific question, you are asking about the situation where $p=4$ and $q=1$.
A: Your question is the same as if we can convert an ARMA(4,1) process to an AR(3) process. Let's check it out!
A stochastic process $\{X_t\}_{t\in \mathbb{Z}}$ is an ARMA(p,q) if:
\begin{equation}
X_t - \phi_1X_{t-1}-\cdots-\phi_pX_{t-p} = Z_t + \theta_1Z_{t-1}+\cdots+\theta_pZ_{t-q}
\end{equation}
where $Z_{t}$ is white noise (WN) with mean 0 and variance $\sigma^2$. An ARMA(4,1) is therefore given by:
\begin{equation}
X_t - \phi_1X_{t-1} - \phi_2X_{t-2}- \phi_3X_{t-3}- \phi_4X_{t-4}=Z_t+ \theta Z_{t-1}
\end{equation}
Considering the lag operator $L^d(Z_t) = Z_{t-d}$, with the condition $|\theta|<1$, we have
\begin{align}
X_t - \phi_1X_{t-1} - \phi_2X_{t-2}- \phi_3X_{t-3}- \phi_4X_{t-4} & = Z_t(1+ \theta L)\\
(X_t - \phi_1X_{t-1} - \phi_2X_{t-2}- \phi_3X_{t-3}- \phi_4X_{t-4})(1+ \theta L)^{-1}  & = Z_t\\
(X_t - \phi_1X_{t-1} - \phi_2X_{t-2}- \phi_3X_{t-3}- \phi_4X_{t-4})\left(1+ \sum_{k = 1}^\infty(-\theta L)^k\right)&=Z_t
\end{align}
with $\lim_{k\rightarrow\infty}\theta^k = 0$, because $|\theta|<1$, the process can be approximated to  an AR(4). For $k=1$ we get an AR(4) - a process with $X_tZ_t$'s. If we don't approximate, then it can never be an AR(something). Note that a MA(1) can be converted to an infinite AR the same as an AR(1) is converted to an infinite MA.
