How to find $\lim_{n\to \infty} \frac{\log(n)}{n}$ without L'Hospital's rule? [duplicate]

$x\leq e^ x$ for all $x\in \mathbb R$.

So $\sqrt n\le e^{\sqrt n}$. Taking log on both sides gives: $\frac 12\log n\le\sqrt n$. It follows that $0\leq \frac 12\frac{\log n}{n}\leq\frac 1{\sqrt n}$. The result follows by Squeeze principle.

Here is another proof, using the integral definition of the logarithm (and the power rule). For $n\geq 1$, we have the following.

\begin{align*} \log n &= \int_1^n \frac1x\ dx\\ &\leq \int_1^n \frac{1}{x^{0.9}}\ dx\\ &=\int_1^n x^{-0.9}\ dx\\[0.5em] &= \frac{x^{0.1}}{0.1}\Big|_{x=1}^{n}\\[0.5em] &= 10n^{0.1} - 10 \end{align*}

Dividing through by $n$, we obtain

$$\frac{\log n}{n} \leq \frac{10}{n^{0.9}} - \frac{10}{n}$$

and the rest follows from the squeeze theorem