Sequence of short exact sequences - does it impose that inverse limits of the terms of the sequence form a short exact sequence?

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I've recently started working on my research project. My supervisor explained to me what exact sequences are and how to construct an inverse limit. I want to apologise for probably not the best language to describe my problem. He asked me a question:

Suppose that we have a sequence of modules and exact sequences: $0\to M_{n}\to N_{n}\to L_{n}\to 0$

Such that for all n:

$p_{n}: M_{n} \to N_{n}$

$q_{n} : N_{n} \to L_{n}$

$p_{n}$ is injective. $q_{n}$ is surjective. And we also have $Im(p_{n}) = ker(q_{n})$.

Does this impose that: $0\to$ inverse limit of $M_{n}\to$ inverse limit of $N_{n}\to$ inverse limit of $L_{n}\to0$?

Are there any pathological situations?

With: p: inverse limit of $M_{n}\to$ inverse limit of $N_{n}$

q: inverse limit of $N_{n}\to$ inverse limit of $L_{n}$

$$\mathrm{Im}(p)=\mathrm{ker}(q)$$

I've proven that if inverse limits "stop changing after some time" (sequences of $M_{n}$, $N_{n}$, $L_{n}$ are finite), then this is true. I did this by assigning $p$ coordinate by coordinate - for example, $p( (x_{1}, x_{2}, \dots, x_{n})) = ( p_{1}(x_{1}),p_{2}(x_{2}), \dots, p_{n}(x_{n}) )$. Analogously with $q_{n}$.

I've been thinking that I might show those pathologies with either a ring with non-zero characteristic, or by chosing one of inverse limits not being closed.

For example: $M_{n}$ = ring of polynomials with variable $x$ over $\mathbb{Q}$ with degree $n$ and $0$th coefficient being equal to $0$, $N_{n} = R^{n+1}$ with $p_{n}(f_{n}) = (f_{n}(0), f_{n}(1), \dots, f_{n}(n+1))$ And $L_{n} = R$ And $q_{n} = \Pi^{n+1}_{1}f(i)$ Then I'd look at the sequence defined as $y_{n} \in M_{n}$ with $y_{n}(x) = \sum_{1}^{n+1}\frac{x^k}{k!}$. Then in the inverse limit of $x$ we'd have $e^x-1$, which is not a rational, creating a "pathology" (???).

I'm looking for hints/tips/ resources where I could figure out this stuff on my own. Please don't use the language of cathergory theory, I'm just a second year bachelor student...