Is $\lim_{x\rightarrow\infty}\frac{f'(x)}{f(x)}=0$ if $\lim_{x\rightarrow\infty}f(x)=0$? 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:


*

*which part of my proof is wrong?

*to make the conclusion true (the limit of such fraction is $0$), is there any additional premise I need?

*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.
 A: 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.
A: 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 $\lim_{x \to \infty} f'(x) \ge c \lim_{x \to \infty} f(x)$.
A: *

*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 $f(x)=e^{-x}$.
A: 
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 $\lim_{x \to \infty} f'(x)/f(x) < 0$.
