Limitm→∞ Limitn→∞ (1+cos2m(n!πx)) where n! denotes factorial of n..
Answer=2 if x is rational and 1 if x is irrational..
1-1 ≤ cos x ≤ 1
0 ≤ cos 2mk ≤ 1
for cos k =1
cos 2m k=1
we have soln = 2
and for other we have cos 2m k →0 therefore limiting value is 1
k=n!x
k should be of the form ,a*pie ;a belongs to N
have you understood
i am not able to explain properly
n not sure of this way
1hint:
if x is rational, it can be written as (1+cos2mn!*pi*(p/q)) where p,q belong to N if not rational, it cannot be expressed as p/q
for each case see what type of a term you get inside the cos, and accordingly decide.
cheers!
1if x is rational
x=pq
and as n→∞
n > q
So, n!q=Integer
cos2(integer*Ï€) =1
hence limit will be 2
In the second case x is irrational
cos2(irrational)<1
lim m→∞cos2m(irrational)=0
hencce limit is 1
