Zeroth right derived functor Let $F:\mathcal{C}\rightarrow \mathcal{D}$ be a left exact functor and $A\in \mathcal{C}$. Assuming $\mathcal{C}$ has enough injectives, we consider an injective resolution $$0\rightarrow A\xrightarrow{\epsilon}I_0\xrightarrow{d_0}I_1\xrightarrow{d_1}I_2\rightarrow \cdots$$
As $F$ is a left exact functor, this induces a complex
$$0\rightarrow F(A)\xrightarrow{F(\epsilon)}F(I_0)\xrightarrow{F(d_0)}F(I_1)\xrightarrow{F(d_1)}F(I_2)\rightarrow \cdots$$
exact at $F(A)$ and $F(I_0)$.
We then define $R^iF(A)=H^i(F(I^*))=\frac{\text{Ker}(F(d_i))}{\text{Im}(F(d_{i-1}))}$
We have $R^0F(A)=\frac{\text{Ker}(F(d_0))}{\text{Im}(F(\epsilon))}$. As above sequence is exact at $F(I_0)$, we have $R^0F(A)=0$. But, in Weibel's Homological Algebra, it is written that, 

since $0\rightarrow F(A)\rightarrow F(I_0)\rightarrow F(I_1)$ is exact, we always have $R^0F(A)=F(A)$.

I do not understand what I am missing. 
 A: This note was also found in Lang's Algebra page 791.  I also found it very confusion.  Hopefully this provides an answer.
$$
0 \to M \xrightarrow{d^{-1}} I^0 \xrightarrow{d^0} I^1 \dots
$$
Applying $F$ a left-exact additive functor, we have:
$$
0 \to F(M) \xrightarrow{F(d^{-1})} F(I^0) \xrightarrow{F(d^0)} F(I^1) \to \dots
$$
$R^0F(M)$ is defined to be $H^0(F(I))$ where $I$ refers to the remainder of the original sequence, with zero added, or $0 \to I^0 \xrightarrow{d^0} I^1 \xrightarrow{d^1} \dots$.
That is $H^0(F(I)) = \ker F(d^0) / \text{im } F(d^{-1}) = \ker F(d^0) / 0$ when we let $f : 0 \to I^0$ is the zero map.  But $\ker F(d^0) = \text{im } F(d^{-1})$.  That is true by exactness of $F$.  But $\text{im } F(d^{-1}) = F(
\text{im } d^{-1}) = F(M)$ since $F$ is an additive functor or in other words an abelian group hom.
So the confusion arises when they all of a sudden remove $M$ from the sequence under consideration and replace it with zero.  I am still confused, and am not sure how / why that it's possible to just throw in a zero.
