Let #L = lim_(x->oo) (1/x)^(1/x)#.
Take the natural logarithm on both sides.
#ln L = lim_(x->oo) 1/x ln(1/x) = lim_(x->oo) 1/x ln(x^-1) =lim_(x->oo)-lnx/x #
Since #lim_(x->oo) lnx/x -> oo/oo#, we can use #color(red)("L'Hôpital's rule")#.
#ln L = - lim_(x->oo) lnx/x = -lim_(x->oo) (d/dx lnx)/(d/dxx)#
#ln L = -lim_(x->oo) (1/x)/1 =-lim_(x->oo) 1/x = -0 = 0#.
Raise #e# to both sides in order to cancel the natural logarithm:
#e^(ln L) = e^0#
#color(red)(L = 1)#
Therefore,
#lim_(x->oo) (1/x)^(1/x) = 1#.
