1. Show lim Θ/sinΘ=1

Θ->0

(Hint: show sinΘcosΘ <Θ<tanΘ)

I have no idea to do this problem, other than use the squeeze theorem, but i still dont know what to do. any help would be greatly appreciated.

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- Oct 4th 2008, 06:33 AMjohntuanlimits and continuity
1. Show lim Θ/sinΘ=1

Θ->0

(Hint: show sinΘcosΘ <Θ<tanΘ)

I have no idea to do this problem, other than use the squeeze theorem, but i still dont know what to do. any help would be greatly appreciated. - Oct 4th 2008, 07:05 AMskeeter
here is a link to the classical geometric approach in determining this limit ...

http://www.csun.edu/~ac53971/courses/math350/xtra_sine.pdf - Oct 4th 2008, 07:32 AMbkarpuz
I will show my own proof for this.

**Proof**.

First draw a unit circle, and put a equilateral $\displaystyle n$-polygon in it.

Then from the center of the circle, draw lines to the corners of the polygon.

We see that the center angle $\displaystyle 2\pi$ is divided to $\displaystyle n$, thus all the triangles have $\displaystyle \frac{2\pi}{n}$ as the vertex of the isosceles triangle, and each triangles have the area $\displaystyle \frac{1}{2}\sin\bigg(\frac{2\pi}{n}\bigg)$.

Hence the sum of the areas of the triangles is $\displaystyle A(n):=\frac{n}{2}\sin\bigg(\frac{2\pi}{n}\bigg)$.

We can see that letting $\displaystyle n$ tend to infinity $\displaystyle A(n)$ tends to the are a of the circle $\displaystyle \pi$.

That is

$\displaystyle \lim\limits_{n\to\infty}A(n)=\pi$

or equivalently

$\displaystyle \lim\limits_{n\to\infty}\frac{n}{2\pi}\sin\bigg(\f rac{2\pi}{n}\bigg)=1$.

Substitute $\displaystyle m=\frac{2\pi}{n}$, then we see that $\displaystyle m\to0$ as $\displaystyle n\to\infty$, which yields

$\displaystyle \lim\limits_{m\to0}\frac{\sin(m)}{m}=1.$ $\displaystyle \rule{0.3cm}{0.3cm}$

I hope this is helpful. (Wink)

**Note**. Here, $\displaystyle m$ travels through rational numbers and since $\displaystyle \lim$ means accumulation, we may think that $\displaystyle m$ travels through reals since it is the closure of rational numbers.