What the limit of (1/x)^1/x as x approaches #oo#? help me please

Question

738 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-30T19:26:53

#lim_(xrarr+oo)(1/x)^(1/x)=1#

#lim_(xrarr+oo)(1/x)^(1/x)# = ?

#(1/x)^(1/x)# #= e^(ln(1/x)^(1/x))# #=e^(1/xln(1/x))# #= e^(1/x(ln1-lnx)# #=# #e^(-lnx/x)# #=# #1/e^(lnx/x)#

so #lim_(xrarr+oo)(1/x)^(1/x)# #=#

#lim_(xrarr+oo)e^(1/xln(1/x)# #=#

#lim_(xrarr+oo)1/e^(lnx/x)# #=A#

#x->+oo#

#u->0#

because #lim_(xrarr+oo)(lnx/x)=lim_(xrarr+oo)1/x=0#
, #lim_(xrarr+oo)lnx=+oo# , #lim_(xrarr+oo)x=+oo#

#y=lnx#
graph{lnx [-10, 10, -5, 5]}

(using Rules De L'Hospital)

As a result, #A=# #lim_(xrarr+oo)1/e^(lnx/x)#

#=lim_(urarr0)1/e^u# #=1/e^0=1#