**EDIT:** 1) edited to make it work for the general case of a short exact sequence 2) added additional assumption following Torsten's comment.

**Assume** in addition that directed colimits are right exact. (Isn't this true always?)

Let 
$0\rightarrow M'\rightarrow M\rightarrow M''\rightarrow0$ be exact with $M'$ and
$M''$ of finite type. Assume $\{P_i\}$ is a directed collection of subobjects of
$M$ such that $\lim P_i=M$.


Let $(M\to Q_i):={\rm coker} (P_i\to M)$ so one has short exact sequences:
$$
0\to P_i \to M \to Q_i \to 0.
$$
Then
$$
\lim P_i = M \quad \Leftrightarrow \quad \lim Q_i =0.
$$ 
Now let $(K'_i\to M'):=\ker (M'\to M\to Q_i)$. Since $\lim Q_i =0$, it follows that $\lim K'_i\to M'$ is surjective and hence there exists a $j$ such that $K'_i\to M'$ is surjective for all $i\geq j$. It follows that $M'\to Q_i$ is the zero map and hence there is an induced map $M''\to Q_i$. Now let $(K''_i\to M''):=\ker (M''\to Q_i)$. Again, since $\lim Q_i =0$, it follows that $\lim K''_i\to M''$ is surjective and hence there exists a $j$ such that $K''_i\to M''$ is surjective for all $i\geq j$ and the map $M''\to Q_i$ is the zero map. So, the original map $M\to Q_i$ is also zero, but then $P_i={\rm ker}(M\to Q_i)=M$.