Question

For each n ∈ N, let fn(x) = (cos x)n. Each fn is a continuous function. Nevertheless, show (a) lim fn(x) = 0 unless x is a multiple of π, (b) lim fn(x) = 1 if x is an even multiple of π, (c) lim fn(x) does not exist if x is an odd multiple of π.

Answer 1

See explanation below.

Explanation:

For this case we have this function:

`f_n *x) = (cos x)^n`

We have that this function is cotinuous and we eant to calculate the
`lim_(n \to \infty) f_n (x) , x\in R`

Part a

From the results above we see that the limit only exists if x is an even multiple of
`\pi`.

For the other case when x is not a multiple of
`\pi` we have that:

`|cos x|<1` and then we can find the limit like this:

`lim_(n \to \infty) f_n (x) = lim_(n \to \infty) (cos x)^n =0`

Because the cos is a number between 0 and 1.

Part b

Assuming that x is an even multiple of
`\pi`, then cos (x)=1.

If x is an even number multiple of
`\pi`.

For example
`x = 2\pi r, r\in Z` we have that we can express:

`cos x = (1)^k`

And on this case
`(cos x)^n = (1)^(kn)`

And for the limit we have that:

`lim_(n \to \infty) f_n(x) =1`.

Part c

Assuming that x is an odd multiple of
`\pi`, then cos (x) =-1

If x is an odd number multiple of
`\pi` for example
`x = \pi (r+1), r\in Z` we have that we can express:

`cos x = (-1)^k`

And on this case
`(cos x)^n = (-1)^(kn)`

And since we have an alternating series we have that this limit:

`lim_(n \to \infty) f_n(x)` not exists.