Double Limits
Evaluate the limit for x ∈ / Q :
n → ∞ lim m → ∞ lim ( cos ( m ! π x ) ) 2 n + 1 ( cos ( m ! π x ) ) 2 n − 1
The answer is -1.
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1 solution
This is not correct. Think about what happens when x = 1 .
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@Abhishek Sinha its given x does not belong to rational numbers set.
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Okay. But how are you applying the limit theorems? Does the limit lim m → ∞ cos 2 n ( m ! π x ) exist?
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@Abhishek Sinha – it does exist and is zero also the power is even
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@Ashutosh Sharma – How do you prove that?
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@Abhishek Sinha – since power is even then value of cosx will lie between 0 and 1 and as n tends to infinity the value of cos^nx tends to zero. sorry i'm not good at latex.
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@Ashutosh Sharma – I am asking about lim m → ∞ cos 2 n ( m ! π x ) , for a fixed n .
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@Abhishek Sinha – for a fixed 'n' which is not a very huge value,value of limit is indefinite but here in this case its given as n tends to infinite in order to calculate limiting value
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@Ashutosh Sharma – That's not rigorous. Can you prove your assertion analytically, from the definition of limits?
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@Abhishek Sinha – He has proved enough. If x is not rational and m is an integer and as cos has even power it would tend to zero . If n was small then it could have been any value from (0,1) but as n is large it tends to 0. I do not even think that it should be a level 4 problem.
just remember that cosx belongs to (-1,1) for all real value let it be infinity too. so if n tends to infinity cos x term tends to zero .thus ans is -1