if a0 = x, a1=x^x, a2= a0^a1 and an= x^a(n-1), if n->infinite equals 2, what is x?

Question

in other words, if x^(x^(x^(...^(x^x) equals to (if it is done to infinite), what is x? I know its sqrt(2), but i dnt know why (I tried to write it as 2^(2^-1), but then?)

Answer 1

a[0] = x
a[n] = x^(a[n - 1])

This means if we take the limit as n approaches infinity of both sides of the recursive question,

lim a[n] = lim x^(a[n - 1])
n -> infinity

But, as given,
lim a[n] = 2
n -> infinity

so our equation becomes

lim x^(a[n - 1]) = 2
n -> infinity

We can extend the limit into the exponent, and note that

lim(a[n]) = lim (a[n - 1]) as n approaches infinity, so it follows that the exponent should equal 2, since
lim a[n - 1] = 2
n -> infinity

x^2 = 2

Take the square root of both sides,

x = +/- sqrt(2)

But, we reject the negative solution because the function is bounded to exclude -sqrt(2), and

x = sqrt(2)

Answer 2

a_(n + 1) = x^a_n

Show that the sequence converges and then

L = lim a_(n + 1) = lim x^a_n = x^(lim a_n) = x^L
But L = 2 so
2 = x^2
x = sqrt2

Answer 3

X is the letter between W and Y.