# limits

• October 4th 2008, 07:33 AM
johntuan
limits 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.
• October 4th 2008, 08:05 AM
skeeter
here is a link to the classical geometric approach in determining this limit ...

http://www.csun.edu/~ac53971/courses/math350/xtra_sine.pdf
• October 4th 2008, 08:32 AM
bkarpuz
Quote:

Originally Posted by johntuan
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.

I will show my own proof for this.
Proof.
First draw a unit circle, and put a equilateral $n$-polygon in it.
Then from the center of the circle, draw lines to the corners of the polygon.
We see that the center angle $2\pi$ is divided to $n$, thus all the triangles have $\frac{2\pi}{n}$ as the vertex of the isosceles triangle, and each triangles have the area $\frac{1}{2}\sin\bigg(\frac{2\pi}{n}\bigg)$.
Hence the sum of the areas of the triangles is $A(n):=\frac{n}{2}\sin\bigg(\frac{2\pi}{n}\bigg)$.
We can see that letting $n$ tend to infinity $A(n)$ tends to the are a of the circle $\pi$.
That is
$\lim\limits_{n\to\infty}A(n)=\pi$
or equivalently
$\lim\limits_{n\to\infty}\frac{n}{2\pi}\sin\bigg(\f rac{2\pi}{n}\bigg)=1$.
Substitute $m=\frac{2\pi}{n}$, then we see that $m\to0$ as $n\to\infty$, which yields
$\lim\limits_{m\to0}\frac{\sin(m)}{m}=1.$ $\rule{0.3cm}{0.3cm}$
I hope this is helpful. (Wink)
Note. Here, $m$ travels through rational numbers and since $\lim$ means accumulation, we may think that $m$ travels through reals since it is the closure of rational numbers.