Using Mean Value Theorem on an example I thought of just using the Mean Value Theorem and plugging in the interval values in it for (a) and (b). Although it seems too easy, am I missing something?
Also how would I do (c)?
Suppose $f:(-\infty,0] \rightarrow \mathbb R$ is continuous everywhere and differentiable on (-$\infty,0$).
Suppose also that $\lim_{x\to -\infty} f(x)=0$.
(a) Show that there exists a $c \in (-1,0)$ such that $f'(c)=f(0)-f(-1).$
(b) Formulate and prove a similar statement in the interval $(-2,-1)$.
(c) Suppose further that $\lim_{x\to -\infty} f'(x)=R$ for some $R \in \mathbb R$. Prove that $R=0$.

By the Mean Value Theorem, for every positive integer $n$ there is a $c_n$ between $-(n+1)$ and $-n$ such that 
$$\frac{f(-n)-f(-(n+1))}{1}=f'(c_n).\tag{1}$$
Since $\lim_{x\to-\infty} f(x)=0$, it follows that the limit of the left-hand side of (1) is $0$, and therefore $\lim_{n\to\infty} f'(c_n)=0$.
Since $\lim_{x\to\infty}=R$, it follows that $R=0$.
Remark: You are right, there is nothing much to (a) and (b), it is all in preparation for (c).
