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.
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.
here is a link to the classical geometric approach in determining this limit ...
http://www.csun.edu/~ac53971/courses/math350/xtra_sine.pdf
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.
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.