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,

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

$$=\lim_{n\to\infty}\left(\dfrac{1}{e^n}\right)^{\ln(3/2)}=0^{\ln(3/2)}=0$$
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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