What Is the Limit of {cos(m!*pi*x)}^{2n} as n and m Approach Infinity?

[anthony.g2013]

Homework Statement

What would be the limit of {cos(m!*pi*x}^{2n} as 'n' and 'm' go to infinity along with proof? Also x is real. I have no clue as to how to start the question. Please help! Thanks.

The Attempt at a Solution

I tried to write the original function as e^[(ln(cos(m!*pi*x)/(1/2n)], but since not both the numerator and denominator go to zero, I couldn't use L'Hopital's Rule.

Other method I tried was using the power series expansion of cosine, but I am not sure how to deal with m!

Any hints or help will be greatly appreciated. Thanks!

 

[vela]

Consider the fact that cosine is bounded between -1 and 1.

 

[anthony.g2013]

Consider the fact that cosine is bounded between -1 and 1.

Actually I should clarify the question and mu doubt:

the actual question is Evaluate lim m→infinity [lim n→ infinity ((cos (m!*pi*x))^2n) ].

So if we went with -1<= cos (m!*pi*x)<= 1, take powers of 2n on all sides and take limit as n goes to infinity, we get the limit of the original function as 1. Is that correct? I am not sure how to deal with the part where m goes to infinity.

 

[vela]

Actually I should clarify the question and mu doubt:

the actual question is Evaluate lim m→infinity [lim n→ infinity ((cos (m!*pi*x))^2n) ].

Are you sure that you're supposed to take the limit as m tends to infinity last? The order will make a difference in how the problem works out.

So if we went with -1<= cos (m!*pi*x)<= 1, take powers of 2n on all sides and take limit as n goes to infinity, we get the limit of the original function as 1. Is that correct? I am not sure how to deal with the part where m goes to infinity.

No, that's not correct. Suppose m=1 and x=1/2. What would be the limit as n→∞? How about if x=1/3 instead?

 

[anthony.g2013]

I see your point. The limit varies as we vary m and x, but I am still stuck. So do we just say that there is no limit because the cosine function keeps oscillating between -1 and 1? I still have no clue how to deal with the m!

 

[vela]

Like I said, the order of the limits matters. When you take the limit as n tends to infinity, you're doing so for a fixed m. And then you take the result of that and see what happens when m goes to infinity.

 

[anthony.g2013]

Could you please show me how you would do the problem? I have tried a variety of things like converting cosine squared into sine squared, or using e^ln form, but nothing seems to work. I am sorry, I am completely blank.

 

[vela]

I see your point. The limit varies as we vary m and x, but I am still stuck. So do we just say that there is no limit because the cosine function keeps oscillating between -1 and 1? I still have no clue how to deal with the m!

I don't think you did actually, because I gave you the wrong examples.

Say m=1. What's the limit equal to when x=1 as n→∞? What's the limit equal to when x=1/3? You should be able to convince yourself there are only two possible outcomes for the inner limit. Which one you get depends on m and x.

 

[anthony.g2013]

Okay let's do it the way you suggest. When m=1, x=1, then cos(pi)^2n=1 as n→∞. When m=1, x=1/3, then cos(pi/3)^2n= (sqrt3/ 2)^infinity = 0. So the two possible outcomes are zero and 1. So since m is an integer, it all comes down to x. If x=integer, then the limit is 1 and if x=/ an integer, then the limit is 0. Am I right yet?

 

[anthony.g2013]

Of course, if m=x, then we will have the limit as 1.

 

[vela]

What if x=1/12? Is the limit always 0 regardless of the value of m?