How to find $\lim\limits_{n\to \infty} \dfrac{\log(n)}{n}$?
It is of no doubt that if we use L'Hospital's rule we will get $\lim\limits_{n\to\infty}\dfrac{ \frac{1}{n}}{1}$ which is of course equal to $0$. But how can we find the limit without using the rule?
I tried to substitute $n = x+1$ so that I could apply exponential series but that also seems to be not working. Is there any other possible method? Or do I have to do another substitution?
On
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
$x\leq e^ x$ 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.