How do you find the limit of #(3/x)^(1/x)# as x approaches infinity?

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Answer 1 · 2016-04-21T20:45:10

#y=1#

#(3/oo)^(1/oo)=0^0#-> this is an indeterminate form so we use l'Hopitals' Rule

Let #y=(3/x)^(1/x)#
#lny=ln(3/x)^(1/x)#

#lny=1/x ln(3/x)#

#lny=ln(3/x)/x#

#lim_(x->oo) lny = lim_(x->oo) ln(3/x)/x #

#lim_(x->oo) lny =lim_(x->oo)((1/(3/x))*-3/x^2)/1#

#lim_(x->oo)lny=lim_(x->oo)(x/3 xx-3/x^2)#

#lim_(x->oo)lny=lim_(x->oo)-1/x#

#lny=-1/oo#

#lny=0#

#e^0=y#

#y=1#