I need to prove that <munder> <mo movablelimits="true" form="prefix">lim <mrow class="MJ

skylsn

skylsn

Answered question

2022-06-24

I need to prove that
lim n 1 n sin ( n π 3 ) = 0.

Answer & Explanation

Braylon Perez

Braylon Perez

Beginner2022-06-25Added 34 answers

The Archimedean property of the real numbers states that for all ϵ > 0 one can always find N N so that 1 N < ϵ. Now, since | sin α | 1 for all α, then it must be the case that
| 1 n sin ( n π 3 ) | 1 n
Now, Let ϵ > 0 be arbitrary. Choose N N so that 1 N < ϵ. Therefore, for all n > N, one has
| 1 n sin ( n π 3 ) | 1 n < 1 N < ϵ
In other words, lim 1 n sin ( n π 3 ) = 0

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?