Tetration Questions: Answers, Calculations & Limitations

[pierce15]

Hello all, of late I have been very interested in tetration (http://en.wikipedia.org/wiki/Tetration). I have a few questions that weren't really explained clearly by this wikipedia article. P.S., I can't get latex to work with the wikipedia notation, so I will use this notation: $x^{\Delta n}$.

1. Why does
$$\lim_{n\to\infty} x^{\Delta n}$$ only converge for $$1/e<x<e^{1/e}$$

2. How can the exact value be calculated as a limit? For example, how could I prove that:
$$\lim_{x\to\infty} \sqrt{2}^{\Delta x}=2$$

3. Why is the following statement true (from the wikipedia article):
$$\lim_{x\to0}\lim_{n\to\infty}x^{\Delta n}=1/e$$

I will be sure to post more questions as they come.

 

[jvicens]

Hello there,

I am glad to hear that you are interested in tetration. It is a fascinating mathematical concept that has been studied for centuries. I will do my best to answer your questions and provide some insight into this topic.

1. The reason why the limit only converges for 1/e<x<e^{1/e} is due to the behavior of the function as x approaches infinity. As you may know, the function x^{\Delta n} grows very rapidly as n increases, and it approaches infinity as x approaches infinity. However, for values of x that are less than 1/e, the function actually decreases as n increases, and it approaches 0 as x approaches 0. This behavior makes it impossible for the limit to converge for those values of x. On the other hand, for values of x that are greater than e^{1/e}, the function grows too quickly and the limit does not exist.

2. The exact value of the limit can be calculated using the definition of a limit. In the example you provided, we can rewrite the expression as \lim_{x\to\infty} (\sqrt{2}^x)^{1/x}. Now, as x approaches infinity, \sqrt{2}^x also approaches infinity, and 1/x approaches 0. This means that the limit can be written as \lim_{x\to\infty} \infty^{0}, which is an indeterminate form. To evaluate this limit, we can use the logarithmic rule for limits: \lim_{x\to\infty} f(x)^{g(x)} = e^{\lim_{x\to\infty} g(x)\ln(f(x))}. Applying this rule to our example, we get \lim_{x\to\infty} e^{0\cdot\ln(\sqrt{2}^x)} = e^0 = 1. Therefore, the exact value of the limit is 1.

3. The statement is true because, as we discussed in the first question, for values of x that are close to 0, the function x^{\Delta n} approaches 1 as n approaches infinity. This means that the inner limit, \lim_{n\to\infty}x^{\Delta n}, is equal to 1 for values of x close to 0. Then, as x approaches 0, the outer limit \lim_{