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

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