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

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Answer 1 · 2018-03-28T06:33:39

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

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

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:

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

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

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