<s>[INST] show that limit $\lim_{n\to+\infty}f_{n}(x)=x+3$,if $f_{n+1}(x)=\sqrt{6(1+x)+f_{n}(x^2)}$ 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
 [/INST] 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.
</s>