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Assume that $f:{\bf R}\to{\bf R}$ is differentiable on ${\bf R}$, and both of $\lim\limits_{x\to\infty}f(x)$ and $\lim\limits_{x\to\infty}f'(x)$ are finite. Geometrically, one may have $$\lim_{x\to\infty}f'(x)=0$$ Here is my question:

How can one actually prove it?

By definition, it suffices to show that

$$\lim_{x\to\infty}\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=0$$

i.e. $$\forall \epsilon>0~\exists M>0\quad \textrm{s.t.}\quad x>M\Rightarrow \left|\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}\right|<\epsilon$$ For large enough $M$ and small enough $\epsilon$, one has $$|f(x+h)-f(x)|<\tilde{\epsilon}$$ But I have no idea how to go on.

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This is a duplicate of math.stackexchange.com/questions/42277/… –  Beni Bogosel Jun 8 '11 at 17:05

3 Answers 3

up vote 9 down vote accepted

You are assuming that the limit of the derivative exists. Say $$\lim_{x\to\infty}f'(x) = L.$$

Case 1. $L\gt 0$. Then there exists $M\gt 0$ such that for all $x\geq M$, $|f'(x) - L|\lt \frac{L}{2}$. Therefore, for all $x\geq M$, $$\frac{L}{2} \lt f'(x) \lt \frac{3L}{2}.$$ In particular, $f$ is increasing on $[M,\infty)$.

By the Mean Value Theorem, for each natural number $n$ there exists a $c$ (which depends on $n$), $M+n \lt c\lt M+n+1$ such that $$f(M+n+1)-f(M+n) = \frac{f'(c)}{(M+n+1)-(M-n)} = f'(c) \geq \frac{L}{2}.$$ Inductively, we conclude that $$f(M+n)\geq f(M)+\frac{Ln}{2}.$$ But as $n\to\infty$, $f(M)+\frac{Ln}{2}\to \infty$; hence $f(x)\to\infty$ as $x\to\infty$, contradicting our assumption that $\lim\limits_{x\to\infty}f(x)$ is finite.

Case 2. $L\lt 0$; a similar argument shows that $f(x)\to-\infty$ as $x\to\infty$ in this case.

Therefore, the only possibility left is $L=0$, as desired.

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The trick to getting geometric information out of the derivative is the mean value theorem. A corollary of the MVT is that, if $m\leq f'(x)\leq M$ for all $x\in [a,b]$, then $m(b-a)\leq f(b)-f(a) \leq M(b-a)$.

So let's proceed to our proof of the the problem.

By subtracting off a constant, we can assume $\lim f(x)=0$. Let $\lim f'(x)=L$. Assume $L\neq 0$. Without loss of generality, $L>0$ Then for $N$ large enough, we have that $x>N$ implies that $|f(x)|<1$ and $L/2<f'(x)<3L/2$.

If $N<a<b$ with $b-a>4/L$, then $f(b)-f(a) > L/2(b-a)> L/2(4/L)=2$, which is impossible.

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This is a relatively old post, but interesting subject, so I thought I will post another solution, also based on the mean value theorem http://en.wikipedia.org/wiki/Mean_value_theorem. Here we go:

$f(x)$ is a primitive for $f'(x)$, using http://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus $\int_{M}^{M+r}f'(x)dx=f(M+r)-f(M)$ for $\forall M\in \mathbb{R}, \forall r>0$

Also (using mean value th), there $\exists \xi (M)\in \left [ M,M+r \right ]$ such that $\int_{M}^{M+r}f'(x)dx=f'(\xi (M))\left ( M+r-M \right )=f'(\xi (M))\cdot r$

So $f(M+r)-f(M)=f'(\xi (M))\cdot r$.

Now, if $M \to \infty $ then $\xi (M) \to \infty $ and $r$ is const. As a result: $0=\lim_{M \to \infty }\left [f(M+r)-f(M) \right ]=\lim_{M \to \infty }f'(\xi (M))\cdot r$ $\Rightarrow \lim_{M \to \infty }f'(\xi (M))=0$

Because $\lim_{x \to \infty }f'(x)$ is finite then it must be 0.

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