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There is a given real function which is differentiable and such that

$$\lim_{x\to\infty}f'(x)=L$$

We're asked to give a proof that if $\;f\;$ is bounded, then the limit must be $\;L=0\;$

What I worked out: geometrically it looks very clear, but I'm having problems with the formal part.

I said: we suppose $\;|f(x)|\le M\;\;\;\forall\ x\in\Bbb R\;$ , and then I tried the mean value theorem. Since

$$\lim_{x\to\infty}\frac{f(x)-f(0)}x=0\;,\;\;\text{because}\;\;|f(x)-f(0)|\le2M\;\;\text{(bounded)}$$

there exists $\;c_x\in(o,x)\;$ for all positive $\;x\;$, so that

$$\frac{f(x)-f(0)}x=f'(c_x)$$

The problem I meet is that if I do$\;x\to\infty\;$ I cannot deduce that $\;c_x\to \infty\;$ because I don't know if we could be getting the same $\;c_x\;$ for different $\;x$'s. If I could solve this then I would get immediately that $\;L=0\;$ .

Any help will be appreciated.

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Sketch of proof. Suppose that $L\ne0$, and without loss of generality that $L>0$. Then there exists $x_0$ such that if $x>x_0$, then $f'(x)>L/2$. Using the mean value theorem, if $x>x_0$ then $$f(x)=f(x_0)+f'(c)(x-x_0)>f(x_0)+\frac L2(x-x_0)\ ,$$ and the RHS is unbounded as $x\to\infty$.

See if you can fill in the details.

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  • $\begingroup$ Thank you very much. It looks really a promise, I will check it. $\endgroup$
    – user177692
    Commented Mar 23, 2015 at 12:36

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