Proving Limit of Squeezed Sequence w/ Squeeze Theorem

[Telemachus]

Homework Statement

I must prove using the squeeze theorem that \displaystyle\lim_{x \to{+}\infty}{n\sin\displaystyle\frac{\pi}{n}}=\pi

Homework Equations

The statement suggests to use \sin(\alpha)\leq{\alpha}\leq{\tg(\alpha)}\in{(0;\pi/2)}

The Attempt at a Solution

I really don't know how to handle it. I've seen how to do this with functions tending to zero, this same kind of trigonometric functions, but I don't know how to work it on a succession like this, tending to infinite.

 

[Mark44]

Homework Statement

I must prove using the squeeze theorem that \displaystyle\lim_{x \to{+}\infty}{n\sin\displaystyle\frac{\pi}{n}}=\pi

Homework Equations

The statement suggests to use \sin(\alpha)\leq{\alpha}\leq{\tg(\alpha)}\in{(0;\pi/2)}

What does (\alpha) mean in the inequality?

What exactly is the problem statement? In particular, does the problem require you to find this limit in a specific way, or are you assuming that you have to do it in this way?

Note that
\lim_{n \to \infty} n sin(\pi /n)~=~ \pi \lim_{n \to \infty} (n/\pi) sin(\pi /n)

The Attempt at a Solution

I really don't know how to handle it. I've seen how to do this with functions tending to zero, this same kind of trigonometric functions, but I don't know how to work it on a succession like this, tending to infinite.

 

[lanedance]

so for the sqeueeze theorem you want to show, for some N, that for all n > N
a_n \leq \pi - sin(\frac{\pi}{n}) \leq b_n

where a_n & b_n both tend to zero with large N

starting with
\sin(\alpha)\leq\alpha} for \alpha \in (0, \pi/2)}
will give you one side of the squeeze

 

[Mark44]

so for the sqeueeze theorem you want to show, for some N, that for all n > N
a_n \leq \pi - sin(\frac{\pi}{n}) \leq b_n

lanedance, you want this as
a_n \leq \pi - n~sin(\frac{\pi}{n}) \leq b_n

where a_n & b_n both tend to zero with large N

starting with
\sin(\alpha)\leq\alpha} [\itex] for \alpha \in (0, \pi/2)}[/tex]<br /> will give you one side of the squeeze

 

[lanedance]

yeah, good pickup cheers - i also have no idea what the (alpha) means either

 

[Telemachus]

Thank you very much for your help. I don't know why it use an alpha, but I think it means that for any number between (0;\pi/2) the \alpha. \pi or whatever would be between the sin of that number, and the tangent of that number. I'll keep working on it, and I'll be back with any news of progress.

 

[Mark44]

You missed my questions.
1. What is the significance of the parentheses in (\alpha)?
2. What is the exact statement of the problem?

 

[Telemachus]

I'm sorry didn't note that, the suggestion was:

\sin(\alpha)\leq{\alpha}\leq{\tan(\alpha)}\in{(0;\pi/2)}

And the statement of the problem says: Prove the limit using the squeeze theorem.

I'm having also other problems with limits, exponential limits of a_n tending to infinite. Always with a_n being a sequence. Can I post some of those limits here?

I couldn't find the way to prove this one neither :(

 

[Mark44]

I see - you omitted "tan" in the OP. Also, you wrote the limit as x increased to infinity, but the expression in the limit involves n. They need to be the same.

The inequality in post #8 is equivalent to 1 <= alpha/sin(alpha) <= 1/cos(alpha) for alpha in (0, pi/2). I got this by dividing all three members by sin(alpha), which is positive on this interval.

This inequality can be rewritten as cos(alpha) <= sin(alpha)/alpha <= 1, taking reciprocals. (If a < b, where a and b are positive, then 1/a > 1/b.)

 

[Telemachus]

Thank you very much Mark.