# Squeezed sequence

1. Apr 20, 2010

### Telemachus

1. The problem statement, all variables and given/known data
I must prove using the squeeze theorem that $$\displaystyle\lim_{x \to{+}\infty}{n\sin\displaystyle\frac{\pi}{n}}=\pi$$

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

3. 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.

2. Apr 21, 2010

### Staff: Mentor

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

3. Apr 21, 2010

### 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

4. Apr 21, 2010

### Staff: Mentor

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

5. Apr 21, 2010

### lanedance

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

6. Apr 21, 2010

### 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.

7. Apr 21, 2010

### Staff: Mentor

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

8. Apr 21, 2010

### 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 :(

9. Apr 21, 2010

### Staff: Mentor

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.)

10. Apr 21, 2010

### Telemachus

Thank you very much Mark.