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Iterative sequences:TIFR 2019 GS Part A, Problem 13

Understand the problem

Suppose we have x_1=\sqrt{2} and we have x_{n+1}=(\sqrt{2})^{x_n}. We have to find out the limit of the sequence.

Source of the problem
TIFR 2019 GS Part A, Problem 13
Topic
Analysis
Difficulty Level
Moderate
Suggested Book
Real analysis, Bartle & Sherbert

Start with hints

Do you really need a hint? Try it first!

I was observing that x_2=(\sqrt{2})^{\sqrt{2}} and x_3=(\sqrt{2})^{{\sqrt{2}}^{\sqrt{2}}} and so on now it is clear that the function is increasing. Can you prove that the sequence is convergent and then how to find the limit?

If x<2, then \sqrt 2^x<2 (the exponential is a growing function). You have a bounded increasing sequence, so it converges. Can you guess where it will converge?
Upon convergence, x=\sqrt2^x, or 2\ln_2 x-x=0. The derivative of the LHS is \dfrac{2}{x\ln2}-1, which has a single root, hence the function has at most two roots. By inspection, they are 2 and 4. Can you get the answer now?
The iterations from x_1=1 do converge to x=2.

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