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

[itex][itex][/itex][/itex]

 

[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.