fn(x) = x[1-x^(2n)] ; x∈[-1,1]
f(x)= lim(n->无穷) fn(x)
case 1: x=-1
f(x)
= lim(n->无穷) x[1-x^(2n)]
=0
case 2: -1
= lim(n->无穷) x[1-x^(2n)]
= x[1-0]
=x
case 3: x=1
f(x)
= lim(n->无穷) x[1-x^(2n)]
=0
ie
f(x)
=0 ; x=-1 or 1
=x ; -1
选第一个答案
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