# 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:

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.

• 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. – Paramanand Singh Jul 2 '17 at 8:43

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.

• thank you so much all, but how about the other questions? – Rizky Reza Fujisaki Jul 1 '17 at 10:10
• 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 $f(x)=e^{-x}$.

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

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

• thank you so much all, I realized my mistake, but how about the other questions? – Rizky Reza Fujisaki Jul 1 '17 at 10:11
• @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. – user335501 Jul 1 '17 at 10:34

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 $\lim_{x \to \infty} f'(x)/f(x) < 0$.