Is it ok to prove that $\,\lim\limits_{n\to\infty}\frac{\log_a(n)}{n}=0\,$ if a > 1 this way? Is it ok to prove that $\,\lim\limits_{n\to\infty}\dfrac{\log_a(n)}{n}=0\,$ this way ?
I tried to prove it using the definition of the limit:
$$\frac{\log_a(n)}{n}<\epsilon.$$
This means that
$$\frac{1}{n}\log_a(n)<\epsilon \iff n^\frac{1}{n}<a^\epsilon.$$
Because $n^\frac{1}{n}\leq n$ if $n$ is a natural number, we can replace $n^\frac{1}{n}$ with $n$ (if we replace $<$ with $\leq$):
$n\leq a^\epsilon$.
Consequently, $N = [a^\epsilon]+1$.
Any mistakes there?
In my textbook it was proved using the squeeze theorem
 A: Your idea is not quite right because, formalizing it, it would be correct if
$$\forall \epsilon>0\ \exists n_0\in\mathbb{N}^*\ \forall n\geq n_0:n\leq a^\epsilon$$
but this is impossible since it would also imply that the natural numbers are bounded by $a^\epsilon$. It is true that $\exists n_0:n_0\leq a^\epsilon$ as you pointed out. But that isn't enough to prove the desired limit since you need that $\forall n\geq n_0:n\leq a^\epsilon$ which is impossible.
The right way of proving it is to notice that your limit is $$\lim_n\log_a(n)/n=\lim_n\log_a(\sqrt[n]{n})=\log_a(\lim_n\sqrt[n]{n})=\log_a(1)=0$$
which makes use of the well-known fact $\lim_n\sqrt[n]{n}=1$ and that $\log_a(x)$ is continuous. However, if you can't use these results, I will show an alternative proof of the said limit.
First, we will prove that $\log(n)/n\to0$ by the squeeze theorem. If $n\geq2$, then
$$n=\left[1+(\sqrt[n]{n}-1)\right]^n=1+\binom{n}{1}(\sqrt[n]{n}-1)+\binom{n}{2}(\sqrt[n]{n}-1)^2+\dots+(\sqrt[n]{n}-1)^n$$
$$n\geq\binom{n}{2}(\sqrt[n]{n}-1)^2\geq0\Rightarrow \frac{2}{n-1}=\frac{n}{\binom{n}{2}}\geq(\sqrt[n]{n}-1)^2\geq0\Rightarrow 1+\sqrt{\frac{2}{n-1}}\geq \sqrt[n]{n}\geq1$$
$$\log\left(1+\sqrt{\frac{2}{n-1}}\right)\geq\frac{\log n}{n}\geq\log(1)=0$$
$$\text{Finally, since }\lim_n\log\left(1+\sqrt{\frac{2}{n-1}}\right)=\log(1)=0, \text{ we have that} \lim_n\frac{\log n}{n}=0$$
Now, noticing that $\log_a(n)=\frac{\log n}{\log a}$, we finally get
$$\boxed{\therefore\lim_n\frac{\log_a n}{n}=\frac{1}{\log a}\lim_n\frac{\log n}{n}=0}$$
