Suppose that $f:[a,\infty)\to \mathbb{R}$ is smooth at infinity. I would like to have some result like if $f$ does not grow too fast, then something like the following holds

(1) $\liminf_{x\to\infty} |f'(x)|/|x f(x)| = 0$.

(2) $\liminf_{x\to\infty} |f'(x)|/|f(x)| = 0$.

I managed to show that (1) holds when $|f(x)|\lesssim e^{x^{1-\delta}}$ for some $\delta > 0$. I think it should hold for $|f(x)|\lesssim e^x$ but I do not have a proof. It actually holds even for $f(x)$ growing faster than $e^x$, say $f(x) = e^{x^{1.1}}$. Does anyone see how to obtain a bigger range of $f$ than $|f(x)|\lesssim e^{x^{1-\delta}}$?

For (2), I can only show that if there exists $\alpha,\beta$ with $\alpha\leq \beta<\alpha+1$ such that $x^\alpha \leq |f(x)| \leq x^\beta$ then (2) holds. Can this be relaxed? I don't feel that a lower bound for $f$ essential?

(Alexandre Eremenko gave the answer to the question I posted earlier. Then I realised the question didn't give what I wanted. Here is the updated question.)

Suppose that $f:[a,\infty)\to \mathbb{R}$ is smooth at infinity. I would like to have some result like if $f'$ is monotone and does not grow too fast, then something like the following holds

(1) $\lim_{x\to\infty} |f'(x)|/|x f(x)| = 0$.

(2) $\lim_{x\to\infty} |f'(x)|/|f(x)| = 0$.

For (2), it is easy to show that if there exists $\alpha,\beta$ with $\alpha\leq \beta<\alpha+1$ such that $x^\alpha \leq |f'(x)| \leq x^\beta$ then (2) holds. Can this be relaxed?

For (1), my feeling is that it should hold for $f'(x)$ up to $f(x) = e^x$ or even up to $e^{x^2}$. But I do not have a proof...