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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
Analysis
Moderate
Suggested Book
Real analysis, Bartle & Sherbert

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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