Question

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=?`

Answer 1

`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 }`