<s>[INST] An interesting limit 
Let $x\in\mathbb{R}.$ For all $i,j\in\mathbb{N},$ define $a_{i0} = \frac{x}{2^i}, a_{ij} = a_{i,j-1}^2 + 2a_{i,j-1}.$ Find, with proof, $\lim\limits_{n\to\infty} a_{nn}.$

Below is my attempt.
Let for each $n, p_n(x) = a_{nn}$. Then observe that $a_{n+1,n} = p_n(\frac{x}2).$ As well, $p_{n+1}(x)+1 = (p_n(\frac{x}2)+1)^2.$ Iterating gives $p_2(x) = ((p_0(\frac{x}4)+1)^2+1)^2, p_3(x) = (((p_0(\frac{x}8)+1)^2+1)^2+1)^2, p_n(x) = ((\dots (p_0(\frac{x}{2^n})+1)^2\dots)^2+1)^2,$ but I'm not sure how this can be converted to a more useful form such as $(1+\frac{x}{2^n})^{2^n}.$
 [/INST] For $n\in \mathbb{N}$ ,
$a_{n,0}=\frac{x}{2^n}$
$a_{n,1}+1=(a_{n,0}+1)^2=(\frac{x}{2^n}+1)^2$
$a_{n,2}+1=(a_{n,1}+1)^2=(\frac{x}{2^n}+1)^4$
Continuing this way,
For $1\le j\le n$,
$a_{n,j}+1=(\frac x{2^n}+1)^{2^j}$
Thus for $n=j$,
$a_{n,n}+1=(\frac x{2^n}+1)^{2^n}$
Taking limit as $n\to \infty$, we get
$lim_{n\to \infty} a_{n,n}+1=e^x$ ( Since the right limit is a subsequence of the sequence $(1+\frac xn)^n$ converging to $e^x$)
This gives required limit as $e^x-1$.
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