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 }