Setup:
I have a tower of abelian groups $\hspace{1em} \cdots \to A_3 \stackrel{f_2}{\to} A_2 \stackrel{f_1}{\to} A_1 \stackrel{f_0}{\to} A_0$.
There are similar towers for $B_i, C_i$, and $D_i$.
There is also an exact sequence $0 \to A_i \xrightarrow{\alpha} B_i \to C_i \to D_i \to 0$.
From this sequence, we derive the short sequence $0 \to \text{coker} \alpha \to C_i \to D_i \to 0$ where $\text{coker} \alpha = B_i/\text{im} \alpha$.
It is known that $\underset{\longleftarrow}{\lim}_i A_i = 0$ and that $\underset{\longleftarrow}{\lim}_i B_i = 0$.
Question:
Does it follow that $\underset{\longleftarrow}{\lim}_i (B_i/\text{im} \alpha) = 0$?
Thoughts:
I proved that $\underset{\longleftarrow}{\lim}_i A_i = 0$ directly from the definition of the inverse limit and that $\underset{\longleftarrow}{\lim}_i B_i = 0$ by showing that the map $\partial$ is injective in the sequence
$$0 \to \underset{\longleftarrow}{\lim}_n B_n \longrightarrow \underset{n}{\prod} B_n \stackrel{\partial}{\longrightarrow} \underset{n}{\prod} B_n \longrightarrow \underset{\longleftarrow}{\lim}^1_n B_n \to 0.$$
Using the second method again, if $\partial$ is injective, then a restriction of the $\partial$ map to $\underset{n}{\prod} B_n/(\text{im} \alpha)$ will also be injective.
This is why I think that $\underset{\longleftarrow}{\lim}_i (B_i/\text{im} \alpha) = 0$. Is this incorrect?
Aside
Obviously I renamed the groups since the original ones are cohomology groups and would take forever to write. I tried to include everything that is necessary for this specific question, but I will edit if I see I'm missing an important detail.
Thank you for your time even reading this!