real analysis - Is $lim_{xrightarrowinfty}frac{f'(x)}{f(x)}=0$ if $lim_{xrightarrowinfty}f(x)=0$?

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:

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.