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

Answer:

See explanation below.

Step-by-step 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:

$|cos x|$ 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.