calculus - Proving $lim_{ntoinfty} left(frac {2}{3}right)^n=0$ using limit definition

Prove $\displaystyle\lim_{n\to\infty} \left(\frac {2}{3}\right)^n=0$ with the definition of limit.

From the definition and since $n\in\mathbb N$ I get that ${\Large\mid} \left(\frac {2}{3}\right)^n{\Large\mid}=\left(\frac {2}{3}\right)^n$ but now I'm not sure what to do, I don't see how taking a $\log$ here would help.

I thought of different approach, proving by induction that $\left(\frac {2}{3}\right)^n < \frac 1 n$: The first steps are trivial, then for $n+1$: $\frac {1}{n+1}>\frac {2}{3}\left(\frac {2}{3}\right)^n$, then from the induction hypothesis: $\frac {1}{n+1}>\frac {2}{3}\frac 1 n$ and we get that this true for $n>2$ so we can choose $N_{\epsilon}=\frac 1 {\epsilon}$.

Is there another way that does not involve induction?

Answer

HINT. By limit definition:
$$|a_n-l|<\epsilon$$
i.e.

$$\left|\left(\frac{2}{3}\right)^n\right|=\left(\frac{2}{3}\right)^n<\epsilon$$
whence
$$-n\ln(3/2)<\ln\epsilon$$
and
$$n>-\frac{\ln\epsilon}{\ln \frac{3}{2}}$$