Proving $ \lim_{n \to \infty} \, \left(\frac{2}{3}\right)^n = 0$ I'm having trouble with a proof of this question:
$$ \lim_{n \to \infty} \, \left(\frac{2}{3}\right)^n = 0$$
I know $N(0)$ is a log of epsilon with base $\frac{2}{3}$ but I can't proceed with a proof.

The limit exists because the sequence is monotonic and bounded.
L'hopital:   $$c:=\lim_{n\to\infty}(\dfrac 23)^n=\lim_{n\to\infty}\dfrac {2^n}{3^n}=\lim_{n\to\infty}\dfrac {\ln2\cdot 2^n}{\ln3\cdot 3^n}=\dfrac {\ln2}{\ln3}\lim_{n\to\infty}\dfrac {2^n}{3^n}=\dfrac {\ln2}{\ln3}\lim_{n\to\infty}(\dfrac 23)^n=\dfrac {\ln2}{\ln3}c$$.
Thus $c=0$.
(You actually don't need L'hopital.   You can just split off a factor of $\frac 23$.)
