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

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?

• $2 log\, (n^{1/2}) \leq 2(n^{1/2}-1)$ Jan 9 at 5:33
• $$n=e^{\log n}>\frac{1}2\log^2 n$$ Jan 9 at 5:39

$$x\leq e^ x$$ 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.

• May I know please, how does $\frac12\log(n) ≤ \sqrt{n}$ follows that $0 ≤ \frac12 \frac{\log(n)}{n} ≤ \frac{1}{\sqrt{n}}$?
– Yooo
Jan 9 at 5:44
• @Utkarsh: log of a natural number is always non negative and hence the first inequality in $0\leq \frac 12\frac{\log n}{n}\color{red}{\leq}\frac 1{\sqrt n}$. The red coloured inequality follows by dividing both sides of $\frac 12\log n\le\sqrt n$ by $n$.
– Koro
Jan 9 at 5:49

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