Let be $f : \mathbb{R}_{+} \to \mathbb{R}$ a continuous and differentiable function.
Let's suppose that $\lim\limits_{x \to +\infty} xf'(x) = 1$. I have to prove that $\lim\limits_{x \to +\infty} f(x) = +\infty$.
What I tried:
- First, we can see that the hypothesis can be rewritten as $f'(x) \sim \dfrac{1}{x}$, we know then $\lim_{x \to +\infty} f'(x) = 0$ and $f'(x) = \dfrac{1}{x} + o\left(\dfrac{1}{x}\right)$.
Idea : Formally, it seems like if I can integrate $f'(x)$, I could get "$f(x) \sim \ln x \to +\infty$" (I know we cannot integrate $o$ asymptotic terms.)
- Second, I'd try to use Taylor formulas, but as $f$ is only derivable, $\int_{0}^{x} f'(t) \textrm{d}t$ is not even defined… (we cannot use any kind of advanced integration such as Lesbegue, etc. Riemann integrals only.)
Hôpital's Rule is unavailable.
The mean value theorem suffices to deduce $f(x) \sim \ln x$.
Given $\varepsilon > 0$, there is an $x_{\varepsilon}$ such that $1-\varepsilon \leqslant xf'(x) \leqslant 1 + \varepsilon$ for $x \geqslant x_{\varepsilon}$. By the mean value theorem we have
$$ f(x+1) - f(x) = f'(\xi_n)$$
for some $\xi_n \in \mathopen]x,x+1\mathclose[$. Then, for $n \geqslant x_{\varepsilon}$ it follows that
$$\frac{1-\varepsilon}{n+1} \leqslant f(n+1) - f(n) \leqslant \frac{1+\varepsilon}{n}$$
and hence
$$(1-\varepsilon)\sum_{k = 1}^m \frac{1}{n+k} \leqslant f(n+m) - f(n) \leqslant (1+\varepsilon)\sum_{k = 0}^{m-1} \frac{1}{n+k}.$$
Thus, using the asymptotics of the harmonic numbers and the eventual monotonicity of $f$, there are constants $K_1,K_2$ such that
$$K_1 + (1-\varepsilon) \ln x \leqslant f(x) \leqslant K_2 + (1+\varepsilon)\ln x$$
for all $x \geqslant x_{\varepsilon}$.