Poorly posed question. You probably mean

whats the limit of (1/x)^(1/x) as x goes to positive infinity, usualy written as

lim (1/x)^(1/x)

x->infinity

With infinity written as an 8 on its side. and the 'x->infinity' part below 'lim'.

A proof requires calculus. The concept of limit is not algebraic.

One can rewrite (1/x)^(1/x) using the following identities:

(1/x) = x^(-1), so

ln (1/x) = - ln x

a^b = e^(b * ln a). Here e = 2.71828... and ln the natural (base e) logarithm and a not zero.

So (1/x)^(1/x) equals

e ^ ((1/x) * (- ln x))

{ provided 1/x is not zero... but we take a limit and we can use a continuity argument to show both expressions have the same limit behavior}

As x goes to infinity 1/x goes to 0, and -ln x goes slowly (slower than 1/x goes to zero) to minus infinity. That is:

lim (1/x) * (-ln x) = 0

x->infinity

{Proof sketch:

(1/x) * (-ln x) = (1 / (ln (e^x))) * (- ln x)) = - (ln x) / (ln (e^x))

and e^x grows much faster than x. Together with the fact that ln is a monotone increasing function on positive real numbers you get the limit above. You may also use

l'Hôpital's rule; see wikipedia link. This works out as follows:

lim (1/x) * (-ln x) = - lim (ln x)/x = - lim (1/x)/1. The last step comes from

lim f(x)/g(x) = lim f'(x)/g'(x), which is l'Hôpital's rule.}

As a result,

lim (1/x) ^ (1/x) = e ^ 0 = 1

x->inf.

QED

Source(s):
http://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule