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Question

Let x be an irrational, then limmlimn{cos(n!πx)}2m equals

A
0
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B
1
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C
1
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D
Indeterminate
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Solution

The correct option is B 0
According to the problem
For n=1,2,3......put
Fn(x)=limx(cosn!πx)2m
when n!x is an integer, Fn(x)=0, For all others values of x,f(x)=limxfn(x)
Hence the irrational X,Fn(x)=0
for every m;
Hence Fn(x)=0, for rational X, say X=pq,
where P and q are interger, we see that
n!x is an interger if mq therefore that
f(x)=1
limmlimn{cos(n!πx)}2m
0 x irrational
1 x rational
Hence the obtained an every where
discontinuous limit function .

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