How can I prove that the $\lim _{n\to \infty }\frac{log_{a}n}{n}=0$ , $a>1$ with the simplest tools of analysis? I tried solving this problem but I didn't know where to start so I searched it on the internet and the only solution that I found was by using L'Hopital's rule. The thing is we haven't studied that in class, but my professor uses this problem constantly in order to solve other problems and I don't understand why this works. I think I should use the squeeze theorem (since I am only familiar with the limits of sequences and that's the most helpful theorem I know), but I don't know which inequality to use. Could you please help me by giving a hint or some advice?
Thanks in advance.
 A: As $\log_a x = \frac{\ln x}{\ln a}$, it is enough to prove the result for $\frac{\ln n}{n}$.
We first prove that $\lim\limits_{n \to \infty} a_n = 0$ where $a_n = \frac{\ln 2^n}{2^n}$. As $\ln 2^n = n \ln 2$, it is enough to prove that $\lim\limits_{n \to \infty} b_n = 0$ where $b_n = \frac{ n}{2^n}$ which is a consequence of the inequality
$$2^n = (1+ 1)^n \ge 1 + n + \frac{n(n-1)}{2}$$ that you can get using binomial theorem.
Finally, you get $\lim\limits_{n \to \infty} \frac{\ln n}{n} =0$ by squeezing $n$ between $2^m$ and $2^{m+1}$
$$\frac{\ln 2^m}{2^{m+1}} = \frac{\ln 2}{2}\frac{m}{2^m} \le \frac{\ln n}{n }\le 2 \ln 2\frac{(m+1)}{2^{m+1}} = \frac{\ln 2^{m+1}}{2^{m}} $$
A: You can use the Squeeze theorem. You probably know that (roughly speaking) when you apply an inverse (continuous) function to a inequality, it "reverts the inequality". So, if $e^x>x$, then $\log(x)<x$. This may be a given property in your notes (I am pretty sure because I saw this in my first real analysis course). Please try it first, and if you cannot, continue reading! :)
$0 \leq \dfrac{\log(n)}{n} = \dfrac{\log(n^{2/2})}{n} = \dfrac{2\log(n^{1/2})}{n} \leq
\dfrac{2\sqrt{n}}{n} = \dfrac{2}{\sqrt{n}} \to 0 $.
So $\dfrac{\log(n)}{n} \to 0$ by the Squeeze theorem.
The difference between $\log(\cdot)$ and $\log_a(\cdot)$ is just a constant, so it is the same proof. Hope this is usefl to you!
A: You know that $\displaystyle \lim_{n\to \infty} n^{1/n}=1$.
$\text{Now}\; \displaystyle \lim_{n\to \infty}{ \log_a n\over n}=\frac{1}{\ln a}\lim_{n\to \infty} {\ln n\over n}={1\over \ln a}\cdot \lim_{n\to \infty} \ln {n^{1/n}}=\frac{1}{\ln a}\cdot \ln \left(\lim_{n\to \infty} n^{1/n}\right)={\ln 1\over \ln a }=0$
