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Question

limmlimn{1+cos2m(n!πx)}.

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Solution

Case 1

If xQ
We know 1cosx10cos2mx1

We can write x=pq, where p,qN

limnn!πpq=kπ, where k is a integer.

cos2mkπ=1

We have limmlimn{1+cos2m(n!πx)}=1+1=2

Case 2

If x is irrational

limmcos2m(k)=0, where k is non integer value.

We have limmlimn{1+cos2m(n!πx)}=1+0=1

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