If x is a real number in [0, 1] then the value oflimn→∞ limm→∞ 1+cos2m⁡(n!πx) is given by

If x∈Q then n!πx will be an integral multiple of πfor large values of n. Therefore, cos⁡(n!πx) will be either 1 or-1 and so cos2m⁡(n!πx)=1limm→∞ limn→∞ 1+cos2m⁡(n!πx)=1∣+1=2If x→Q, n!πx will not be an integral multiple of π and socos⁡(n!πx) will lie between -1 and 1thus, limm→∞ cos2m⁡(n!πx)=0⇒ limm→∞ limn→∞ 1+cos2m⁡(n!πx)=1+0=1.Thus, option (a) is correct.