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let $x$ is give postive real number,if $f_{0}(x)=0,0<x\le\dfrac{1}{2}$, and such $$f_{n+1}(x)=\sqrt{6(1+x)+f_{n}(x^2)}$$ show that $$\lim_{n\to+\infty}f_{n}(x)=x+3$$ This problem is from AMM 11967(2017),this solution

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  • there is no canonical reason why they chose the functions the way they did and came up with inequality they did. Usually, one solves a long, hard problem in a messy way but has to write the solution neatly. No one in the history of the world just looks at this problem and immediately writes down the function used. – dezdichado May 13 '20 at 00:40
  • There is a typo in the AMS solution. It should be $4$ on place of $1/4$. – Mittens May 13 '20 at 01:02

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Let $g_n(x) = f_n(x) - x - 3$, then $g_0(x) = -(x + 3)$ and it suffices to prove that $$ \lim_{n\to \infty} |g_n(x)| = 0. $$

Note that since $$ \begin{align} g_{n+1}(x) = f_{n+1}(x) - (x + 3) &= \sqrt{6(1+x) + x^2 + 3 + g_n(x^2)} - (x + 3)\\ &=\frac{g_n(x^2)}{\sqrt{(x+3)^2 + g_n(x^2)} + (x + 3)}, \end{align} $$ we have $$ |g_{n+1}(x)| < \frac13|g_{n}(x^2)|. $$ Therefore, $$ 0 \leq |g_n(x)| < \frac13 |g_{n-1}(x^2)| < \dots < \frac{1}{3^n} |g_0(x^{2^n})| = \frac{|x^{2^n} + 3|}{3^n}, $$ and the result immediately follows from the squeeze theorem.

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