Determine limit without l'Hopital

Question

I've found some other pretty similar questions, but I have to find the limit:

$$\lim_{n\to \infty} \frac{\log n}{n} $$

using the fact that $$\log n = \int_1^n \frac{1}{x} dx $$

and $x^{-1}\leq x^{-\frac{1}{2}}$ if $x>1$.

Finding the limit in some way by l'Hopital or comparison, etc. isn't that difficult, however, we have to use the integral part since we have to use is in other questions.

Can you give me a hint?

Answer 1

$$if,n>1 \\\log(n) \leq \sqrt{n}\\ $$so $$0\leq\lim_{n \to \infty }\frac{logn}{n}<\lim_{n \to \infty } \frac{\sqrt{n}}{n}\rightarrow 0$$

Answer 2

Let $ a>0$, then for $n\ge 1$ you have

$$0\le \frac {\ln n}{n^a}=\frac{\int_1^n\frac {dx}{x}}{n^a}\le \frac{\int_1^n\frac {dx}{x^{1-a/2}}}{n^a} = \frac 2a \frac{ {n^{a/2}-1}}{n^a}\to0\quad\text{as}\quad {n\to\infty}$$

Answer 3

Let $x=\frac{1}{n}$. Then, $$\lim_\limits{n\to +\infty}{x}=0^+$$ and $$\lim_\limits{n\to +\infty}{\frac{\log n}{n}}=\lim_\limits{x\to 0^+}{(x\cdot \log \frac{1}{x})}=-\lim_\limits{x\to 0^+}{(x\cdot \log x)}=0$$