Power functions and $\lim_{x \to +\infty}{\frac{\int_{0}^{x}{f(t)\mathrm{d}t}}{x f(x)}}=\frac{1}{1 + \beta}$ Let $f : \mathbb{R}_+ \rightarrow \mathbb{R}_+$ an increasing continuous function such that $f(0) = 0$ and $\beta \geq 0$. When we have for all $x > 0$, $\displaystyle\frac{\int_{0}^{x}{f(t)\mathrm{d}t}}{x f(x)} = \frac{1}{1 + \beta}$, it is straightforward that $f$ is a power function, i.e. there is $\lambda > 0$ such that for any $x \geq 0$, $f(x) = \lambda x^{\beta}$.

My question is:
Does $\lim\limits_{x \rightarrow +\infty}{\displaystyle\frac{\int_{0}^{x}{f(t)\mathrm{d}t}}{x f(x)}} = \displaystyle\frac{1}{1 + \beta}$ imply that there is some $\lambda \geq 0$ such that $\lim\limits_{x \rightarrow +\infty}{\displaystyle\frac{f(x)}{x^{\beta}}} = \lambda$ ?

 A: In general, $f$ will be “almost” a power function. More precisely, it will be regularly varying with index $\beta$, i.e., of the form $f(x)=x^\beta\cdot\ell(x)$ where $\ell$ is a slowly varying function, such as $\ell(x):=\log(1+x)$. This is in essence the content of Karamata's theorem (see Theorems 1.5.11 and 1.6.1 of Bingham, Goldie, and Teugel's book “Regular variation”): Karamata's theorem states that a positive, locally bounded function $f:[0,\infty)\to[0,\infty)$  is a regularly varying function with index $\beta\ge0$ if and only if
$$\frac{x^{\sigma+1}f(x)}{\int_0^xt^\sigma\,f(t)\,\mathrm dt}\xrightarrow[x\to\infty]{}\sigma+\beta+1\tag{$C_\sigma$}$$
holds for some $\sigma>-(\beta+1)$. In this case, $(C_\sigma)$ holds for all $\sigma\ge-(\beta+1)$.
The representation theorem (Theorem 1.3.1 in the book) says that we can write any slowly varying function $\ell$ in the form $$\ell(x)=c(x)\exp\left(\int_0^x\varepsilon(u)\frac{\mathrm du}u\right)$$
for some $c$ measurable with $c(x)\to c\in(0,\infty)$ and $\varepsilon(x)\to0$ as $x\to\infty$.
