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

This is quiet a simple question, but still I'm not sure that I am correct.

Let $f$ be a differentiable function in $\mathbb{R}$ such that:

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

We need to prove that there is $a\in \mathbb{R}$ such that $f'(a)=0$.

If f is permanent, it is obvious.

If it is not, there must be a maximum or minimum.

I know (Maybe I am wrong) that If the limits in $\infty$ and $-\infty$ are final, so the function is uniformly continuous by Kantor theorem, and then we may say that If it uniformly continuous it is bounded, and thus gets it minimum and maximum? (By weierstrass theorem?), and Finally we use Rolle' theorem.

Am I correct? Did I wake up the calculus masters for nothing? How would you answer this question?

Thank you

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or perhaps it means that it is bounded and that's it?! – user6163 Jun 16 '11 at 9:06
If it has final limits in addition in infinity and minus infinity, so I believe it is bounded – user6163 Jun 16 '11 at 9:16
Nir: +1 for showing your work, I sincerely appreciate it. Never mind the recent kerfuffle :) – t.b. Jun 16 '11 at 9:24
Thank you Theo :-) – user6163 Jun 16 '11 at 9:37

## 2 Answers

Since this is trivial if $f$ is identically zero, suppose it is not. Thus $f(r) \neq 0$ for some $r \in \mathbb{R}$. Since $f$ is continuous and $\lim _{x \to \infty } f(x) = \lim _{x \to -\infty } f(x) = 0$, there exist $r_1 < r$ and $r_2 > r$ such that $f(r_1)=f(r_2)$. Hence, since $f$ is everywhere differentiable, there exists some $a$ between $r_1$ and $r_2$ such that $f'(a)=0$.

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Do you need some elaboration? – Shai Covo Jun 16 '11 at 9:46
No, it's cool, wasn't me that voted down. – user6163 Jun 16 '11 at 9:55
@Nir, thank you. – Shai Covo Jun 16 '11 at 10:01

If $f$ is not constant, the exists $x_0$ such that $f(x_0)\neq 0$. We can assume that $f(x_0)>0$ (if it's not the case we will consider $-f$). We use the definition of the limit for $\varepsilon =\frac{f(x_0)}2$. We can find $x_1$ and $x_2$ with $x_1<0<x_2$ such that $|f(x_1)|< \frac{f(x_0)}2$ and $|f(x_2)|<\frac{f(x_0)}2$. Hence we have $f(x_1)< \frac{f(x_0)}2$ and $f(x_2)< \frac{f(x_0)}2$. By the intermediate value theorem we can find $y_1$ and $y_2$ with $x_1<y_1<x_0<y_2<x_2$ such that $f(y_1)=f(y_2)=\frac{f(x_0)}2$. Now we conclude by Rolle's theorem.

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Your answer is fine (except that "mean value theorem" should be replaced with "intermediate value theorem"), hence +1 from me. – Shai Covo Jun 16 '11 at 11:57