2
$\begingroup$

I have just solved a problem:

Let $f:[0,+\infty)\rightarrow \mathbb{R}$ be continuous on $[0,+\infty)$ and differentiable on $(0,+\infty)$. If $\displaystyle \lim_{x\rightarrow +\infty}f'(x)=0$, prove that $\displaystyle \lim_{x\rightarrow +\infty}\frac{f(x)}{x}=0$

My questions is about the inverse: if $\displaystyle \lim_{x\rightarrow +\infty}\frac{f(x)}{x}=0$, which hypothesis can be added (if needed) so that we can conclude $\displaystyle \lim_{x\rightarrow +\infty}f'(x)=0$ ?

Actually, I tried to solve the following one, but I still cannot solve it:

If $f:[0,+\infty)\rightarrow [0,+\infty)$ such that its second derivative is continuous, $f'\le 0$ and $|f''|\le M$ for some $M$ for all $x\ge 0$, then $\displaystyle \lim_{x\rightarrow +\infty}f'(x)=0$.

From the hypothesis, $f$ decreases and bounded below, so $f$ has a limit when $x$ tends to infinity. Thus $\displaystyle \lim_{x\rightarrow +\infty}\frac{f(x)}{x}=0$.

My questions seems to be in a wider range than the second problem above. Any help would be appreciated.

$\endgroup$
  • $\begingroup$ Possibly realated $\endgroup$ – Winther Apr 4 '15 at 16:35
  • $\begingroup$ For the 1st problem (which you have already solved) L'Hospital's is the shortest solution. Its converse is not true in general because of the simple reason that a function may be bounded without the derivative being bounded. For the second problem note that $f$ is non-negative and decreasing and hence has a finite limit $L$ as $x \to \infty$. Now the solution is available at math.stackexchange.com/q/1196439/72031 $\endgroup$ – Paramanand Singh Apr 4 '15 at 22:05
1
$\begingroup$

The second statement you list is true; in particular, it follows from the fact that bounding the second derivative above means that if $f'(x)$ drops below some $-\varepsilon$, then there must be a corresponding drop in the value of $f(x)$, since the derivative cannot return to $0$ arbitrarily quickly.

In particular, suppose, for some $x$, that $f'(x)<-\varepsilon$. It follows that if we take a function $g$ such that $g(x)=f(x)$ and $g'(x)=-\varepsilon$ but $g''(x)=M$, then $f(y)\leq g(y)$ for all $y\geq x$. However, $g(y)$ can be explicitly found as: $$g(x)-g(x+t)=\varepsilon t - \frac{M}2t^2$$ and so $g(x)-g(x+t)$ has a maximum at $t=\frac{\varepsilon}{M}$ with a value of $\frac{\varepsilon^2}{2M}$. Thus, if $f'(x)<-\varepsilon$, the function $f$ is bound to drop by at least $\frac{\varepsilon^2}{2M}$. However, given that $\lim_{x\rightarrow\infty}f(x)=c$ exists and is a lower bound for $f$, it follows that if $c\leq f(x)<c+\frac{\varepsilon^2}{2m}$, which is true for all large enough $x$, then $f'(x)\geq -\varepsilon$. One can easily from here establish that $\lim_{x\rightarrow\infty}f'(x)=0$.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.