3
$\begingroup$

Let $f$ is continuously differentiable positive function, $\lim_{x\rightarrow\infty}f(x)=0$, and $\lim_{x\rightarrow\infty}f'(x)$ exist, is it true that $\lim_{x\rightarrow\infty}\frac{f'(x)}{f(x)}=0$.

My answer was yes, and this is my "proof".

Since the limit of $f'(x)$ exists, we can conclude that $\lim_{x\rightarrow\infty}f'(x)=0$. Suppose that $\lim_{x\rightarrow\infty}\frac{f'(x)}{f(x)}>0$. Hence, there are $c,k\in\mathbb{R}^+$ such that for every $x>k$, we have $\frac{f'(x)}{f(x)}>c$, or $f'(x)>cf(x)$. By letting $x\rightarrow\infty$, we shall get $0>0$, which is a contradiction.

However, my Sensei told me that this proof is wrong, but he didn't tell me which part. He just gave me a counter example that if $f=\phi$ (pdf of standard normal distribution), then we shall get the limit is infinity.

So, my questions are:

  1. which part of my proof is wrong?
  2. to make the conclusion true (the limit of such fraction is $0$), is there any additional premise I need?
  3. how about $\lim_{x\rightarrow\infty}\frac{[f'(x)]^2}{F(x)}$, where $F$ is the antiderivative of $f$? (I think if I can solve the problem of the second question, I'll get some ideas for the third)

Thank you so much in advance.

$\endgroup$
  • $\begingroup$ There is one fundamental problem apart from those mentioned in various answers. How do you know that limit of $f'/f$ exists? If numerator and denominator both tends to $0$ then the ratio can show all kinds of limiting behavior including oscillation and infinities. $\endgroup$ – Paramanand Singh Jul 2 '17 at 8:43
3
$\begingroup$

There is a problem when you write that “By letting $x\to\infty$, we shall get $0>0$”. No! The only conclusion that you reach here is that $0\geqslant0$. Therefore, there is no contradiction.

$\endgroup$
  • $\begingroup$ thank you so much all, but how about the other questions? $\endgroup$ – Rizky Reza Fujisaki Jul 1 '17 at 10:10
3
$\begingroup$
  • With stronger hypothesis (like $f>0$), your problem is equivalent to $$\lim_{x\to \infty }g(x)=-\infty \implies \lim_{x\to \infty }g'(x)=0,$$ which is of course wrong.

  • But with simple intuition, if $\lim_{x\to \infty }f'(x)$ exist and is not $0$, then it's bounded at $+\infty $, and since $\lim_{x\to \infty }f(x)=0$, you either have that $\lim_{x\to \infty }\frac{f'(x)}{f(x)}=\pm\infty $ or doesn't exist.

  • For a counter example, take $f(x)=e^{-x}$.

$\endgroup$
  • 1
    $\begingroup$ Presumably you're making the substitution $g(x) = \log f(x)$; it's worth at least mentioning that! $\endgroup$ – user14972 Jul 1 '17 at 9:42
3
$\begingroup$

Your conclusion $f'(x) > c f(x) $ implies $0 \gt 0$ (when $x \to \infty$ is wrong.

If $f'(x) > cf(x) $ this implies that when we take $x \to \infty$ then $\lim_{x \to \infty} f'(x) \ge c \lim_{x \to \infty} f(x)$.

$\endgroup$
  • $\begingroup$ thank you so much all, I realized my mistake, but how about the other questions? $\endgroup$ – Rizky Reza Fujisaki Jul 1 '17 at 10:11
  • $\begingroup$ @RizkyRezaFujisaki you can take $f(x) = e \ ^ {-x}$ , so $f>0$ , $lim f(x) = 0$ . but $lim \dfrac{f'(x)}{f(x)} = lim \dfrac{- e \ ^ {-x}}{e \ ^ {-x}} = lim -1 = -1 \ne 0$ so the statement is false. $\endgroup$ – user335501 Jul 1 '17 at 10:34
0
$\begingroup$

which part of my proof is wrong?

Let's plug in the counterexample you were given and find out. Letting $f(x) = \exp(-x^2/2)/\sqrt{2\pi}$,

  • $ \lim_{x\rightarrow\infty}f'(x)=0 $
  • $ \lim_{x \rightarrow \infty} f'(x)/f(x) = -\infty$

So we already see a gap in your proof: you completely forgot to consider the case that $\lim_{x \to \infty} f'(x)/f(x) < 0$.

$\endgroup$

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.