How can I effectively prove that f(L) = L in this problem?

[DivGradCurl]

Problem:

(a) Let a_1 = a, a_2 = f(a), a_3 = f(a_2) = f(f(a)), \ldots, a_{n+1} = f(a_n), where f is a continuous function. If \lim _{n \to \infty} = L, show that f(L) = L.

(b) Illustrate part (a) by taking f(x) = \cos x , a = 1, and estimating the value of L to five decimal places.

My answer:

(a) \lim _{n \to \infty} a_{n+1} = \lim _{n \to \infty} f(a_n) = f \left( \lim _{n \to \infty} a_n \right) = f(L) = L

(b) I have used my calculator to get this one. First, I plugged in: \cos 1. I took the cosine of the result. Then, I kept on taking the cosine until my results agreed to five decimal places. I got L \approx 0.73908.

My question:

Did I get it right? Are there other ways to find answer (b)?

Thanks a lot!

 

[Gokul43201]

0.73908 looks right assuming a=1 rad.

 

[phoenixthoth]

Problem:

(a) Let a_1 = a, a_2 = f(a), a_3 = f(a_2) = f(f(a)), \ldots, a_{n+1} = f(a_n), where f is a continuous function. If \lim _{n \to \infty} = L, show that f(L) = L.

(b) Illustrate part (a) by taking f(x) = \cos x , a = 1, and estimating the value of L to five decimal places.

My answer:

(a) \lim _{n \to \infty} a_{n+1} = \lim _{n \to \infty} f(a_n) = f \left( \lim _{n \to \infty} a_n \right) = f(L) = L

(b) I have used my calculator to get this one. First, I plugged in: \cos 1. I took the cosine of the result. Then, I kept on taking the cosine until my results agreed to five decimal places. I got L \approx 0.73908.

My question:

Did I get it right? Are there other ways to find answer (b)?

Thanks a lot!

I think it would be more convincing if you wrote L= first on the left hand side and then came to f(L) on the right hand side. You basically have f(L)=L in your argument which is what you want to prove. Right idea though.