67th Romanian Mathematical Olympiad

(i)⇒(ii). Let $\epsilon >0$; according to (i), there exists $m_1>0$ such that $\frac{f(x)}{x^{a+\epsilon }}<1,\forall x>m_1$ and $m_2>0$ such that $\frac{f(x)}{x^{a-\epsilon }}>1,\forall x>m_2$. If we denote $m=\max \{m_1,m_2,1\}$, then $x^{a-\epsilon }<f(x)<x^{a+\epsilon },\forall x>m$. This implies $a-\epsilon <\frac{\ln f(x)}{\ln x}<a+\epsilon ,\forall x>m$. Since $\epsilon >0$ is arbitrarily chosen, it follows that ${\lim }_{x\to \infty }\frac{\ln f(x)}{\ln x}=a$.

(ii)⇒(i). Let $\epsilon >0$. From (ii), we derive the existence of some $m>1$ such that $$a-\frac{\epsilon }{2}<\frac{\ln f(x)}{\ln x}<a+\frac{\epsilon }{2},\forall x>m.$$ This yields $x^{a-\epsilon /2}<f(x)<x^{a+\epsilon /2},\forall x>m$, and therefore $\frac{f(x)}{x^{a+\epsilon }}<\frac{1}{x^{\epsilon /2}},$ for all $x>m$, and $\frac{f(x)}{x^{a-\epsilon }}>x^{\epsilon /2},\forall x>m$. But ${\lim }_{x\to \infty }{x}^{\epsilon /2}=\infty$ and $f(x)>0,\forall x>0$, from which we obtain the desired conclusion.