Limit of $n/\ln(n)$ without L'Hôpital's rule I am trying to calculate the following limit without L'Hôpital's rule:

$$\lim_{n \to \infty} \dfrac{n}{\ln(n)}$$

I tried every trick I know but nothing works. You don't have to prove it by definition.
 A: Let $\ln(n) = t$. We then have $n = e^t$. Hence, we have
$$\lim_{n \to \infty} \dfrac{n}{\ln(n)} = \lim_{t \to \infty} \dfrac{e^t}{t}$$
Recall that $e^t = 1 + t + \dfrac{t^2}{2!} + \cdots > \dfrac{t^2}{2}$. Hence, we have
$$\lim_{t \to \infty} \dfrac{e^t}{t} > \lim_{t \to \infty} \dfrac{t^2}{2t} = \lim_{t \to \infty} \dfrac{t}{2} = \infty$$
A: Every time you make $n$ twice and big, you increase $\ln n$ by less than $1$ (since $e$ is less than $2$).  So imagine repeatedly doubling the numerator and each time you do it, adding just $1$ to the denominator, and think about what that will approach.  That does it.
PS inspired by a comment:
$$
\frac{n}{\ln n} > \frac{n}{\log_2 n}.
$$
A comparison test then says that if the latter goes to $\infty$, then so does the former.
Each value of $n$ is located between two powers of $2$: we have $2^m\le n\le 2^{m+1}$. That number $m$ is $m=\lfloor \log_2 n\rfloor$.  So $m\le\log_2 n<m+1$.
$$
\frac{2^m}{m+1}\le\frac{n}{m+1}<\frac{n}{\log_2 n}\le\frac n m.
$$
So it is enough to show that $\dfrac{2^m}{m+1}\to\infty$.  Show by induction that for some constant $c$, we have
$$
\frac{2^m}{m+1} \ge \left(\frac 3 2\right)^m
$$
and then do another comparison.
