Question

cos ((2n+1)pi/2) as limits n approaches infinity

Answer 1

It's a cosine. Why would it approach a limit in that direction? I'm pretty sure is continue to oscillate between 1 and -1.

Answer 2

so the limit doesnt exist?

Answer 3

Why are you asking? If it oscillates, it doesn't have a limit. Go with confidence.

Answer 4

If your expression continues to alternate between -1 and 1, then no, no limit exists.
Why not actually evaluate this expression for n=1, 2, 3, 4, 5 and see what you get?
It'd only take a moment.

\[\cos ((2n+1)\pi/2).\]

Answer 5

If n=1, we'll have cos 3Pi/2. Please evaluate that before you do anything else.

Answer 6

Mathmale is on correct, plug in n=1 and see what you get. The limit will exist.

Answer 7

Try your best to explain why that limit will exist, @Mankaran.

Answer 8

#1 "plug in" shouldn't exist. Never do that. Never say that. Yes, I've seen it in textbooks, too. It shouldn't be there. There is no such operation and substitution is absolutely NOT the thing to do unless you have established continuity. Is anyone thinking about continuity, here?

#2 It is correct that Integer Values for n produce odd integer multiples of \(\pi/2\), creating the illusion that the limit, as stated, takes on ONLY the value zero. However, this is not correct, AS STATED.

#3 It is often assumed that "n" takes on only integer values. However, this is NOT in the problem statement. Therefore, the assumption is erroneous, given present information.

The question that remains, "Should n take on only Whole Number values?" It's not in the presented problem statement. Maybe the OP just forgot to mention it? It may be stated at the beginning of the problem section or perhaps implicitly in the course materials or classroom discussion.

The correct answer, as presented, is that the limit does not exist.

The correct answer, assuming \(n \in \mathbb{N}\), is the expression takes on ONLY the value zero (0), so the limit of the sequence is zero (0). Last I checked, zero (0) exists.