limit of n sin (pi/n) as n --> infinity? This question has been asked in:

Does the sequence converge or diverge? $an+ n[\sin(\pi/n)]$

That post/title is messy to understand, so please refer to the problem as stated in my title.

I don't understand the answers and am unable to comment, so I had to re-ask the question here. In the first answer, it is stated:

"It diverges, but for a different reason. sinx &lt; x for x∈(0,∞)"
Why is this true?

It then states, "sinx≥0 for x∈(0,π/2]"
I understand this is true, but do not see how this is applied once the inequality is set up for the squeeze theorem.



My own personal work is as follows:

-1 &lt;= sin (pi/n) &lt;= 1
-n &lt;= n sin(pi/n) &lt;= n
using the range of sin(x), then multiplying by n.

The limits of both sides as n--> inf are not the same, so I'm unable to proceed from here.

Thanks!

NOTE:
The textbook answer is given as "pi."
Rewrite as $$\lim_{n\rightarrow \infty}\pi \frac{\sin(\pi/n)}{\pi/n}$$

We know that $\lim_{x\rightarrow 0}\frac{\sin(x)}{x} = 1$ so that is enough to determine that the limit exists.
We will use the following 2 facts:


For $x&lt;10^\circ, \sin x\approx x$
As $n\to\infty$, $\frac{\pi}{n}\to0$


Fact 2 implies that, for large $n$, $\sin \frac{\pi}{n} \approx \frac{\pi}{n}$.  Then
$$\lim_{n\to\infty} (n\cdot\sin(\frac{\pi}{n})) = \lim_{n\to\infty} (n\cdot\frac{\pi}{n}) = \lim_{n\to\infty} \pi = \pi$$