$\lim\limits_{x\to\infty}xf'\left(x\right)=1\Rightarrow\lim\limits_{x\to\infty}f\left(x\right)=\infty$ 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.
 A: 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.

A: We prove under that general case. That is, we just assume that $f'$
exists on $(0,\infty)$.
Choose $x_{0}$ such that $xf'(x)>\frac{1}{2}$ whenever $x\geq x_{0}$.
For any positive integer $n>x_{0}$, by Mean-Value Theorem, we have
that
\begin{eqnarray*}
f(n+1)-f(n) & = & f'(\xi_{n})\\
 & \geq & \frac{1}{2\xi_{n}}\\
 & \geq & \frac{1}{2(n+1)}.
\end{eqnarray*}
where $\xi_{n}\in(n,n+1).$ Fix an integer $n_{0}>x_{0}$. For any
$k\in\mathbb{N}$, we have that
\begin{eqnarray*}
f(n_{0}+k)-f(n_{0}) & = & \left[f(n_{0}+k)-f(n_{0}+k-1)\right]+\left[f(n_{0}+k-1)-f(n_{0}+k-2)\right]+\cdots+\left[f(n_{0}+1)-f(n_{0})\right]\\
 & \geq & \frac{1}{2}\left\{ \frac{1}{n_{0}+k}+\frac{1}{n_{0}+k-1}+\cdots+\frac{1}{n_{0}+1}\right\} .
\end{eqnarray*}
This shows that $f(n_{0}+k)\rightarrow\infty$ as $k\rightarrow\infty$
because $\sum_{k=1}^{\infty}\frac{1}{n_{0}+k}=\infty$.
Let $M>0$ be given. Choose $k_{0}$ such that $f(n_{0}+k)>M$ whenever
$k\geq k_{0}$. Let $x_{1}=n_{0}+k_{0}$. For any $x>x_{1}$, we have
that $f(x)>f(x_{1})>M$ (because $f'(x)>\frac{1}{2x}>0\Rightarrow f$
is strictly increasing on $[x_{1},\infty)$. ). Hence, $\lim_{x\rightarrow\infty}f(x)=\infty$.
A: We prove under the extra assumption that $f'$ is Riemann integrable over any interval $[a,b]$, for $a>0$.
Since $\lim_{x\rightarrow\infty}xf'(x)=1>\frac{1}{2},$ there exists
$x_{0}$ such that $xf'(x)>\frac{1}{2}$ whenever $x\geq x_{0}$.
For $x>x_{0}$, we have that
\begin{eqnarray*}
f(x)-f(x_{0}) & = & \int_{x_{0}}^{x}f'(t)dt\\
 & \geq & \int_{x_{0}}^{x}\frac{1}{2t}dt\\
 & = & \frac{1}{2}\left(\ln x-\ln x_{0}\right).
\end{eqnarray*}
That is, $f(x)\geq f(x_{0})+\frac{1}{2}\left(\ln x-\ln x_{0}\right).$
It follows that $\lim_{x\rightarrow\infty}f(x)=\infty$ because $\lim_{x\rightarrow\infty}\ln x=\infty.$
