I learned at school that the fact that sinx/x goes 1 when x approaches 0 can be proved by comparing areas.
I'm wondering if there is another way to prove it. Do you have any idea about it?
Asked
Active
Viewed 817 times
1
-
In fact, this can be taken as an axiom, along with the angle addition formulas and even/oddness for $\sin$ and $\cos$. These 5 equations can be used to prove all the other properties (if I remember correctly). – mr_e_man Apr 30 '19 at 05:19
-
1Needed would be some kind of definitions for $\sin x$, and for $x$ in radians]. – coffeemath Apr 30 '19 at 05:19
-
2Can you tell us what you dislike about the argument with areas that leads you to seek other proofs? Will you dislike arguments based on arc length comparisons too? It is essential that radian measure is used for sin(x), so in some way that feature should make an appearance. (Using power series seems like overkill.) – KCd Apr 30 '19 at 05:20
-
I define sinx and cosx by the unit-circke definition. – Soling Apr 30 '19 at 05:22
-
1Which "unit-circle" definition? – Angina Seng Apr 30 '19 at 05:23
-
I'm just a curious to know if there is other geometrical ways to prove it without areas – Soling Apr 30 '19 at 05:23
-
I meant unit-circle definition by https://en.m.wikipedia.org/wiki/Sine – Soling Apr 30 '19 at 05:23
2 Answers
3
In the diagram, $x$ is the signed length of the arc $DB$, and $\sin x$ is the signed length of the segment $AB$ and $\tan x$ is the signed length of the segment $DC$.
Either $0<\sin x<x<\tan x$ or $\tan x<x<\sin x<0$.
In either case, $0<\dfrac{\sin x}{x}<1<\dfrac{\tan x}{x}=\dfrac{\sin x}{x}\cdot\dfrac{1}{\cos x}$.
Thus $\cos x<\dfrac{\sin x}{x}<1$.
Since $\lim_{x\to0}\cos x=1$ we have $\lim_{x\to0}\dfrac{\sin x}{x}=1$ by the "squeezing" theorem.
ADDENDUM Why is $x<\tan x$ when $0<x<\frac{\pi}{2}$?
Consider the following diagram:
Construct the tangent to the unit circle at $B$ and mark its intersection with the tangent line $CD$ as $E$. Then $E$ is equidistant from both $D$ and $B$. Construct the circle with center $E$ and radius $BE$. Then we have $x=\text{ arc }DB<BE+ED<DC=\tan x$ See link
Note that $C$ lies on a tangent line to the smaller circle and is different from the point of tangency, thus it lies outside the smaller circle.
John Wayland Bales
- 21,814
-
-
That's a fair question. Rather than repeat what has already been done, however, there are several good answers already on math.stackexchange. – John Wayland Bales Apr 30 '19 at 06:45
-
1Ok, I think my proof is better than the ones given in the link, so I went ahead and added it. – John Wayland Bales Apr 30 '19 at 07:10
-
-
@AnshumanAgrawal Good catch! It certainly cannot be because of $\theta/2<\tan(\theta/2)$. I will see if I can find a better reason. – John Wayland Bales Feb 02 '22 at 17:54
-
@AnshumanAgrawal I have posed this as a math.stackexchange question. – John Wayland Bales Feb 02 '22 at 20:12
-
Thanks, I was confused and thought I was missing something obvious – Anshuman Agrawal Feb 03 '22 at 05:55
-
I also read this answer: https://math.stackexchange.com/a/922970/930811 But couldn't follow the argument for why that holds, in the comments people were saying that's essentially an axiom – Anshuman Agrawal Feb 03 '22 at 06:35
0
There can be another simple intuitive proof for understanding purpose only. Consider a triangle with one angle x at O. And hypotenuse 1. Then the perpendicular gives the magnitude of sin x. Now just imagine the angle x is very small. Then what, the perpendicular would be vary small and the base is nearly the length of hypotenuse. Now to approximate you can consider perpendicular to be a small arc of circle, with radius as the base(=which is same as hypotenuse of unit length), centre at O. Now, using the definition of angle, we can write: x= (arc length/radius)= sin x. Hence, the limit sin x/x tends to 1 for very small x. Hope this helps you in some way!
Tojra
- 1,483
- 7
- 13