TITLE: $\lim\limits_{x\to\infty}xf'\left(x\right)=1\Rightarrow\lim\limits_{x\to\infty}f\left(x\right)=\infty$
QUESTION [7 upvotes]: Let $f:\left(0,\infty\right)\longrightarrow\mathbb{R}$ differentiable on $\left(0,\infty\right)$ and $\lim\limits_{x\to\infty}xf'\left(x\right)=1$.
Show that: $\lim\limits_{x\to\infty}f\left(x\right)=\infty$.
I get that for a sufficently large x: $\frac{1}{2x}<f'\left(x\right)<\frac{3}{2x}$. But I'm not sure how to proceed.

REPLY [6 votes]: By L'Hopital rule we have
$$\lim_{x\to+\infty} \frac{f(x)}{\ln x} = \lim_{x\to+\infty} \frac{f'(x)}{\frac1x} = \lim_{x\to+\infty} xf'(x) = 1$$
so
$$\lim_{x\to+\infty} f(x) = \lim_{x\to+\infty} \left(\frac{f(x)}{\ln x}\cdot \ln x\right) = \left(\lim_{x\to+\infty} \frac{f(x)}{\ln x} \right)\left( \lim_{x\to+\infty}\ln x\right) = 1 \cdot (+\infty) = +\infty.$$
As @Sangchul Lee pointed out in the comments, this is valid by Theorem 5.13. in Rudin since $\lim_{x\to+\infty} \ln x = +\infty$. Paraphrased, the theorem says:

Let $-\infty \le a < b \le +\infty$ and let $f,g : (a,b) \to \Bbb{R}$ be two differentiable functions such that $g'(x)\ne 0$ for all $x \in (a,b)$. Suppose
$$\lim_{x\to b} \frac{f'(x)}{g'(x)} = A$$
for some $-\infty \le A \le +\infty$.
If $\lim_{x\to b} g(x) = +\infty$, then
$$\lim_{x\to b} \frac{f(x)}{g(x)} = A$$
as well.