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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 $R=0$.

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  • $\begingroup$ Part a doesn't make sense; $f$ isn't defined at $0$ $\endgroup$ – Michael Harrison Mar 3 '16 at 4:24
  • $\begingroup$ @MichaelHarrison A typo, fixed it, thanks for pointing out. $\endgroup$ – NeoXx Mar 3 '16 at 4:28
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    $\begingroup$ Ah, okay. In that case, you are right, it's a direct application of MVT. For part c, is it intuitively clear? $\endgroup$ – Michael Harrison Mar 3 '16 at 4:31
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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).

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HINT:

$$f'(\xi)=\frac{f(-n)-f(-n-1)}{-n-(-n-1)}=f(-n)-f(-n-1)$$

and $\lim_{x\to -\infty}f(x)=0$

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