MHB Limit of a Geometric Sequence: How to Evaluate?

[tmt1]

I have this limit:

$$\lim_{{n}\to{\infty}} {(\frac{2}{3})}^{n}$$

I know the answer is 0 but how can I evaluate this?

 

[Greg]

Can you agree that $\lim_{n\to\infty}\dfrac{1}{e^n}=0$, without calculation?

If so,

$$\lim_{n\to\infty}\left(\dfrac23\right)^n=\lim_{n\to\infty}e^{n\ln(2/3)}=\lim_{n\to\infty}e^{-n\ln(3/2)}$$

$$=\lim_{n\to\infty}\left(\dfrac{1}{e^n}\right)^{\ln(3/2)}=0^{\ln(3/2)}=0$$