proof that lim[x-->0] sin(x) / x=1.Something about it I d

[xailer]

hello

The following is proof that lim[x-->0]sin(x)/x = 1

We draw the unit circle and observe that if

A is the origin,
C is a point on the unit circle in the first quadrant
C' is the projection of C on the x axis
B is the point (1, 0)
B' is the point on the line AC such that the project of B' onto the x axis is B
x is the angle CAB

then

sin x is the length of segment CC'
x is the length of the arc CB
tan x is the length of segment BB'

and

sin x <= x <= tan x

and therefore

1 / sin x >= 1 / x >= 1 / tan x

sin x / sin x >= sin x / x >= sin x / tan x

1 >= sin x / x >= cos x

as x -> 0, cos x -> 1, therefore sin x / x also approaches 1.

I fail to understand how this picture of unit circle with triangles in it proves that tan(x) is bigger than x ( x is angle in radians ). From the drawing of unit circle we can see that sin(x) < x and it makes total sense. But I have no clue looking at the same drawing of unit circle why tan(x) would be greater than arc CB.
Of course in the drawing the line BB'( representing tan(x) ) appears to be greater than arc CB. But that is because the triangle ABB' is bigger than ACC'.We could very well draw a new circle with radius BB' and then line line BB' ( tan(x) ) would be smaller than x.

If I'd knew why we can conclude from the drawing that tan(x) > x, I'd understand the above proof.

thank you

 

[jolly]

Have you tried to draw it yet? http://www.math2.org/mmb/thread/35853

 

[pka]

The lengths \(\displaystyle \L
|OA| = |OB| = 1\) because of the unit circle.

Now the area in yellow is \(\displaystyle \L
(1/2)|OD||OB|\sin (x) = (1/2)|OD| \sin (x)\).

The area of the circular sector, yellow+green, is \(\displaystyle \L
(1/2)x\).

The area of larger triangle, yellow+green+blue, is \(\displaystyle \L
(1/2)|OA||AC| = (1/2)|AC| = (1/2)\tan (x)\).

Comparing areas you see that: \(\displaystyle \L
(1/2)|OD|\sin (x) < (1/2)x < (1/2)\tan (x)\)

Divide and get \(\displaystyle \L
|OD| < \frac{x}{{\sin (x)}} < \frac{1}{{\cos (x)}}\).

As \(\displaystyle \L
x \to 0\quad \Rightarrow \quad |OD| \to 1\quad \& \quad \frac{1}{{\cos (x)}} \to 1\).

That is the usual proof.

 

[xailer]

I understand most of the proof but for one thing!

The lengths \(\displaystyle \L
|OA| = |OB| = 1\) because of the unit circle..

NOTE:[Even though you used different letters for angles I will use the original ones]



I understand that. I also know that for arc x to represent angle in radians, circle must have radius of 1.
I understand that when it comes to unit circle and sin(x). But I don't understand it when it comes to tan(x). Why would BB' represent actual tan(x)? By that I mean why would BB'/arc x actually be the correct ratio of tan(x)/x?

Wouldn't be more correct ( when trying to find tan(x) ) if we say AC' ( where C' is the projection of C on the x axis ) is 1 and then CC' would represent tan(x) ( I know it's wrong since in that case sin(x) = tan(x) ) ?

 

[pka]

Well, because I can’t see your diagram I have no idea what you’re asking!
Please post the diagram.

 

[xailer]

I'm really lost at words, since I have no idea how ask the thing that's bothering me!

Well, because I can’t see your diagram I have no idea what you’re asking!
Please post the diagram.

uh I don't know how to draw one. I will try to describe it using your diagram.
On your diagram we have unit circle and in it inscribed ODB triangle.
Ratio of DB/arc(x) = sin(x)/x.

But triangle OAC seems contrived. AC does actually represents (if |OA|=1)
ratio AC/OA = tan(x). But why does DB/AC represent actual ratio sin(x)/tan(x) and why should we based on your diagram conclude (even though I know it is true) that arc(x) < tan(x)? As I've said in my previous post, I know why the size of arc(x) in the diagram is actually in right proportion compared to sin(x), but I don't know why AC aka tan(x) on the diagram is of right proportions compared to arc(x) and DB aka sin(x).

 

[pka]

Please go to the top of this webpage.
You will see a tab labeled as “Forum Help”.
The first item on that tab is “Inserting Images”.
That will help you insert your diagram.

Your description is very clear to you. I am sure of that!
But I cannot follow it with any surety.

 

[pka]

But triangle OAC seems contrived. AC does actually represents (if |OA|=1) ratio AC/OA = tan(x). But why does DB/AC represent actual ratio sin(x)/tan(x) and why should we based on your diagram conclude (even though I know it is true) that arc(x) < tan(x)? As I've said in my previous post, I know why the size of arc(x) in the diagram is actually in right proportion compared to sin(x), but I don't know why AC aka tan(x) on the diagram is of right proportions compared to arc(x) and DB aka sin(x).

I have a suggestion for you!
Go to your library. There should be several Calculus textbooks there.
If you are not in a college, go to a local college library.
Every calculus textbook in a library will have this theorem.
If possible find the text A FIRST COURSE IN CALCULUS by Lang.

 

[xailer]

Ha!I get it now. It was right under my nose the whole time. And if I wouldn't complicate the matter so much I'd see it sooner

thank you all for your kind help