I'd love your help with finding the following

Jasiah Vincent

Jasiah Vincent

Answered question

2022-04-06

I'd love your help with finding the following limit:
limncos(πn2n)

Answer & Explanation

firenzesunzc65

firenzesunzc65

Beginner2022-04-07Added 16 answers

cos(πn2n)=(1)ncos(π(n2nn))
=(1)ncosπnn2n+n
=(1)ncosπ111n+1
hence |cos(πn2n)|=|cos(π111n+1)|. Since limn+π111n+1=π2, the cos  is continuous and cosπ2=0 we conclude that the limit is 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?