Limit of $\underset{n\to \infty }{\text{lim}}\frac{\ln (n+1)}{\ln (n)}$ without L'Hôpital I intuitively understand that the limit goes to 1 and I can solve with L'Hôpital but I can't without it.
I tried to call it equal to y and rise both sides to the base e but doesn't seems to work.
$\underset{n\to \infty }{\text{lim}}\frac{\ln (n+1)}{\ln (n)}$

You have
$$\frac{\ln (n+1)}{\ln n} = \frac{\ln n(n+1/n)}{\ln n} = 1+  \frac{\ln (1+1/n)}{\ln n}$$ and $\lim\limits_{n \to \infty}\ln (1+1/n) = 0$.
