Calculus: Early Transcendentals, Chapter 4, 4.4, Section 4.4, Problem 61

Replacing oo for x in limit equation yields the nedetermination oo^o . You need to use the logarithm technique, such that:
f(x) = (x)^(1/x)
You need to take logarithms both sides, such that:
ln f(x) = ln (x)^(1/x))
Using the property of logarithms yields:
ln f(x) = (1/x)*ln x
ln f(x) = (ln x)/x
You need to evaluate the limit:
lim_(x->oo) ln f(x) = lim_(x->oo) (ln x)/x = oo/oo
You need to use L'Hospital theorem:
lim_(x->oo) (ln x)/x = lim_(x->oo) ((ln x)')/(x')
lim_(x->oo) ((ln x)')/(x') = lim_(x->oo) (1/x)/1
lim_(x->oo) 1/x = 1/oo = 0
Hence, lim_(x->oo) ln f(x) = 0 , such that lim_(x->oo) f(x) = e^0 = 1
Hence, evaluating the given limit, using l'Hospital rule and logarithm technique yields lim_(x->oo) (x)^(1/x) = e^0 = 1 .