If #A_x=(-1/(x^2+1),1/(|x|+1)]# how do you find ? #uuu_(x=1)^(oo)A_x= ?# and #nnn_(x=1)^(oo)A_x=?#

1 Answer
Mar 28, 2018

#uuu_(x=1)^oo A_x = A_1 = (-1/2, 1/2]#

#nnn_(x=1)^oo A_x = { 0 }#

Explanation:

Given:

#A_x = (-1/(x^2+1), 1/(abs(x)+1)]#

Note that:

  • #-1/(x^2+1) < 0# for all #x in RR#

  • #lim_(x->oo) -1/(x^2+1) = 0#

  • #1/(abs(x)+1) > 0# for all #x in RR#

  • If #0 < a < b# then:

    #-1/(a^2+1) < -1/(b^2+1) < 0 < 1/(abs(b)+1) < 1/(abs(a)+1)#

  • #A_1 = (-1/2, 1/2]#

So:

#(-1/2, 1/2] = A_1 sup A_2 sup A_3 sup ... sup { 0 }#

and for any #epsilon > 0# there is some #n in NN# such that:

#epsilon !in A_x ^^ -epsilon !in A_x# for all #x >= n#

So:

#uuu_(x=1)^oo A_x = A_1 = (-1/2, 1/2]#

#nnn_(x=1)^oo A_x = { 0 }#