MHB Limit of a Geometric Sequence: How to Evaluate?

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The limit of the geometric sequence is evaluated as follows: the limit of (2/3)^n as n approaches infinity is indeed 0. This is supported by the fact that the exponential function e^n grows faster than any polynomial or geometric sequence with a base less than 1. The transformation of (2/3)^n into an exponential form using natural logarithms confirms that it approaches zero. Thus, the limit can be concluded without extensive calculations. The final result is that the limit is 0.
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I have this limit:

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

I know the answer is 0 but how can I evaluate this?
 
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Can you agree that $\lim_{n\to\infty}\dfrac{1}{e^n}=0$, without calculation?

If so,

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

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

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