What is $x$ if $\lim_{n\to\infty}{a_n} = 2$ while $a_1=x$ and $a_{n+1}=x^{a_n}$

I was told to find $x$ if I know ($a_n$)'s limit while $a_1=x$ and $a_{n+1}=x^a_n$, specifically for $\lim_{n\to\infty}{a_n}=2$ and then for different limits. It is pretty clear that $x=\sqrt{2}$, but I can't for the world of me think of a way to prove it (by the way, we were never taught a statement saying that if $\lim_{n\to\infty}{a_n} = L$ then $\lim_{n\to\infty}{x^{a_n}} = x^L$, I'm not sure if it's true, but if it is I think I have to prove it).

Any help will be greatly appreciated! Thanks.

• I am certain this is a duplicate. – Najib Idrissi Nov 27 '13 at 22:51

The statement you mention is true, and it is expressed by saying that "exponentiation is a continuous operation". You can show it by noting that, given a base, there exists a $\delta>0$ that will make $B^{x\pm \delta}$ within epsilon of $B^x$. Knowing this, simply take the limit of both sides of the expression $$a_{n+1}= x^{a_n}$$