$(i)\implies (ii)$ Let $\epsilon >0$. From the epsilon-delta definition of the limit we have that $\exists n_1, n_2 >0$ such that $\frac{f(x)}{x^{a+\epsilon}}<1,\forall x>n_1$ and $\frac{f(x)}{x^{a-\epsilon}}>1,\forall x> n_2$. Let $N=\max\{n_1,n_2\}$. We have that $$x^{a-\epsilon} <f(x)<x^{a+\epsilon},\forall x>n \implies a-\epsilon < \frac{\ln(f(x))}{\ln x}< a+\epsilon,\forall x>n.$$Since $\epsilon$ was chosen arbitrarily, the conclusion follows. $(ii) \implies (i)$ Let $\epsilon >0$. Again, from the epsilon-delta definition of the limt we have that $\exists N>0$ such that $$a-\frac{\epsilon}{2} <\frac{\ln(f(x))}{\ln x} < a+ \frac{\epsilon}{2},\forall x> N.$$Hence, $x^{a-\frac{\epsilon}{2}}<f(x) < x^{a+\frac{\epsilon}{2}},\forall x>N $. From here it follows that $$0<\frac{f(x)}{x^{a+\epsilon}}<\frac{1}{x^{\frac{\epsilon}{2}}},\forall x>N \implies \lim_{x\to \infty}\frac{f(x)}{x^{a+\epsilon}}=0$$and that $$\frac{f(x)}{x^{a-\epsilon}}>x^{\frac{\epsilon}{2}},\forall x>N \implies \lim_{x\to \infty}\frac{f(x)}{x^{a-\epsilon}}=\infty $$