sinx/x and the Squeeze Theorem

Aug 2010
961
101
The standard geometric derivation for derivative of sinx depends on lim sinx/x, which in turn is based on the squeeze theorem. This broadly based and accepted misconception is the reason for this post.

Aside: You can’t define derivative of sinx from the series because you need derivative to get the series (OK, you can define sinx by the series, but then you have to show that it’s the same as the geometric definition, unless you want to abandon all geometrical meaning, in which case you have to derive trigonometric identities from series.)

The squeeze theorem: If
1) g(x) ≤ f(x) ≤ h(x) and lim g(x) = lim h(x) = L, Then Lim f(x) = L

From the geometric definition you get:
2) cosx < sinx/x < 1

It is then pretty universally concluded (Thomas Calculus, Wicki, google, etc, etc etc) that lim sinx/x = 1 by the Squeeze Theorem which is clearly incorrect since the conditions of the squeeze theorem are not satisfied.
The correct conclusion is that lim sinx/x = 1 by definition of limit, since by 2) sinx/x can be made arbitrarily close to 1.

In an effort to satisfy the requirements of the squeeze theorem many authors (I don’t know why they are so hung up on this) reach the following from the geometry: (google sinx/x and squeeze theorem)
cosx ≤ sinx/x ≤ 1
Which is obviously incorrect from the geometry.
 
Jun 2009
675
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The usual geometric approach leads to \(\displaystyle \frac{1}{\cos \theta}>\frac{\theta}{\sin\theta}>1\) doesn't it ?

The Squeeze Theorem works perfectly well with that.
 

Prove It

MHF Helper
Aug 2008
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unit circle.JPG

This is a unit circle, obviously with a radius of 1 unit. The angle $\displaystyle \begin{align*} \theta \end{align*}$ is the angle swept out from the positive x axis in the anticlockwise direction, measured in radians. The green length is $\displaystyle \begin{align*} \cos{(\theta)} \end{align*}$, the red length is $\displaystyle \begin{align*} \sin{(\theta)} \end{align*}$ and the purple length is $\displaystyle \begin{align*} \tan{(\theta)} \end{align*}$. From the diagram, it's clear that the area of the segment is a little bigger than the area of the smaller triangle, and a little less than the area of the large triangle.

We'll deal with positive angles for the moment:

The area of the small triangle is $\displaystyle \begin{align*} \frac{1}{2}\sin{(\theta)}\cos{(\theta)} \end{align*}$, the area of the large triangle is $\displaystyle \begin{align*} \frac{1}{2}\tan{(\theta)} = \frac{1}{2}\cdot \frac{\sin{(\theta)}}{\cos{(\theta)}} \end{align*}$, and the area of the segment is $\displaystyle \begin{align*} \frac{\theta}{2\pi} \cdot \pi \cdot 1^2 = \frac{1}{2}\theta \end{align*}$, thus...

$\displaystyle \begin{align*} \frac{1}{2}\sin{(\theta)}\cos{(\theta)} \leq \frac{1}{2}\theta &\leq \frac{1}{2}\cdot \frac{\sin{(\theta)}}{\cos{(\theta)}} \\ \sin{(\theta)}\cos{(\theta)} \leq \theta &\leq \frac{\sin{(\theta)}}{\cos{(\theta)}} \\ \cos{(\theta)} \leq \frac{\theta}{\sin{(\theta)}} &\leq \frac{1}{\cos{(\theta)}} \\ \frac{1}{\cos{(\theta )}} \geq \frac{\sin{(\theta)}}{\theta} &\geq \cos{(\theta ) } \\ \cos{(\theta )} \leq \frac{\sin{(\theta)}}{\theta} &\leq \frac{1}{\cos{(\theta)} } \end{align*}$

And now as $\displaystyle \begin{align*} \theta \to 0, \cos{(\theta)} \to 1 \end{align*}$ and $\displaystyle \begin{align*} \frac{1}{\cos{(\theta)}} \to 1 \end{align*}$, thus $\displaystyle \begin{align*} \frac{\sin{(\theta)}}{\theta} \to 1 \end{align*}$ as $\displaystyle \begin{align*} \theta \to 0 \end{align*}$.


Of course this only proves the right hand limit, where you are approaching 0 from positive values of $\displaystyle \begin{align*} \theta \end{align*}$, you need to prove the left hand limit as well, where you make $\displaystyle \begin{align*} \theta \to 0 \end{align*}$ from negative values as well. But the proof of that is almost identical :)
 
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Aug 2010
961
101
In response to previous 2 posts:

1) The squeeze theorem is:
If g(x) ≤ f(x) ≤ h(x) and lim g(x) = lim h(x) = L, Then Lim f(x) = L

2) I am familiar with geometric definition of lim sinx/x.
It is correctly expressed with < rather than <= because chord length can never equal arc length.

There is a difference between a has to be less than b, and a can be less than or equal to b.
 

Prove It

MHF Helper
Aug 2008
12,897
5,001
In response to previous 2 posts:

1) The squeeze theorem is:
If g(x) ≤ f(x) ≤ h(x) and lim g(x) = lim h(x) = L, Then Lim f(x) = L

2) I am familiar with geometric definition of lim sinx/x.
It is correctly expressed with < rather than <= because chord length can never equal arc length.

There is a difference between a has to be less than b, and a can be less than or equal to b.
There is nothing wrong with what I wrote, when x = 0 the area of all three shapes is 0, and so their areas would be equal. I never said ANYTHING about chord or arc lengths.
 
Aug 2010
961
101
Look at your own figure. It is impossible for areas to be = because it is impossible for the vertical sides to equal arc length. You can define them equal in the limit but that is what you are trying to prove, which you can't by the squeeze theorem. And that's the point.
 
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Prove It

MHF Helper
Aug 2008
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Sure they can, when the heights of the triangles and the arclength on the segment are all equal to 0.
 
Aug 2010
961
101
The inequality only applies when x =' 0, in which case you can only use <. You can't use what you are trying to prove in your hypothesis.

=': unequal

EDIT, Sinx/x is not defined for x=0, so you can't use it in your hypothesis.

Perhaps you should revisit the squeeze theorem above.

There is a difference between f(x) has a property for a<x<b and a<=x<=b.
 
Last edited:

Prove It

MHF Helper
Aug 2008
12,897
5,001
Surely you can see that when x = 0, $\displaystyle \begin{align*} \frac{1}{2}\sin{(x)}\cos{(x)} = \frac{1}{2}\sin{(0)}\cos{(0)} = 0 \end{align*}$, $\displaystyle \begin{align*} \frac{1}{2}x = \frac{1}{2}(0) = 0 \end{align*}$ and $\displaystyle \begin{align*} \frac{\sin{(x)}}{2\cos{(x)}} = \frac{\sin{(0)}}{2\cos{(0)}} = 0 \end{align*}$. These quantities are all valid! They are all equal!
 
Aug 2010
961
101
Surely you can see that when x = 0, $\displaystyle \begin{align*} \frac{1}{2}\sin{(x)}\cos{(x)} = \frac{1}{2}\sin{(0)}\cos{(0)} = 0 \end{align*}$, $\displaystyle \begin{align*} \frac{1}{2}x = \frac{1}{2}(0) = 0 \end{align*}$ and $\displaystyle \begin{align*} \frac{\sin{(x)}}{2\cos{(x)}} = \frac{\sin{(0)}}{2\cos{(0)}} = 0 \end{align*}$. These quantities are all valid! They are all equal!
Yes, but sinx/x is not defined for x=0. I note that wiki gets squeeze theorem and inequality right (only<), but draws the wrong conclusion, (Lim = 1 by squeeze theorem) the subject of OP.