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
Explanation:
Given:
A_x = (-1/(x^2+1), 1/(abs(x)+1)]
Note that:
-
-1/(x^2+1) < 0 for allx in RR -
lim_(x->oo) -1/(x^2+1) = 0 -
1/(abs(x)+1) > 0 for allx 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 !in A_x ^^ -epsilon !in A_x for allx >= n
So:
uuu_(x=1)^oo A_x = A_1 = (-1/2, 1/2]
nnn_(x=1)^oo A_x = { 0 }