The usual geometric approach leads to doesn't it ?
The Squeeze Theorem works perfectly well with that.
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.
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
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.
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.
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.
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!
I suspect your knowledge of limits is limited. A function does NOT have to be defined at a point in order for it to have a limit there.
In fact, if it WAS the case that a function had to be defined at a point in order to have a limit at that point, there would be no such thing as a derivative, since the derivative is defined as $\displaystyle \begin{align*} \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \end{align*}$, even though the function is NOT defined at h = 0.
As another example, consider the limit $\displaystyle \begin{align*} \lim_{x \to 0}\frac{x^2 + x}{x} \end{align*}$. If you drew a graph of the function, you would see that it is identical to $\displaystyle \begin{align*} x + 1 \end{align*}$, except that it has a hole where x = 0. But the function still approaches the same value as you close in on the point x = 0 from both directions. You would most likely have been taught to factorise and cancel, and then substitute the value in. This is fine, in fact it's exactly what you are supposed to do, and you would find that the limiting value is 1. But the function is NOT defined there.
More specifically, the precise definition of a limit is that if you can show for all $\displaystyle \begin{align*} \epsilon > 0 \end{align*}$ that there exists a $\displaystyle \begin{align*} \delta > 0 \end{align*}$ such that $\displaystyle \begin{align*} 0 < \left| x - c \right| < \delta \implies \left| f(x) - L \right| < \epsilon \end{align*}$ then it proves $\displaystyle \begin{align*} \lim_{x \to c} f(x) = L \end{align*}$. Notice the fact that $\displaystyle \begin{align*} 0 < \left| x - c \right| \end{align*}$, this literally means that there has to be some distance between x and c, which means that $\displaystyle \begin{align*} x \neq c \end{align*}$, thereby enabling the function to not need to be defined at $\displaystyle \begin{align*} x = c \end{align*}$ in order for there to be a limit there.
[QUOTE=Prove It;822408] A function does NOT have to be defined at a point in order for it to have a limit there. /QUOTE]
Correct, that is why I wonder that you keep using sinx/x, x=0 (incorrect) in your hypothesis.
Perhaps it would help to see it done correctly, not using the squeeze theorem (incorrectly):
Just out of curiosity, I thought I’d look for someone who gets it right, not that easy. Courant, Dif and Int Calc vol 1 pg 48, does:
Using the same figure you do, but a slightly different lower area based on chord instead of the triangle vertical (which is tighter), he correctly gets:
“1 < x/sinx < 1/cosx”
And correctly concludes:
“We know that cos x tends to 1 as x→∞ and from this it follows that the quotient sinx/x can differ only arbitrarily little from 1, provided that x is near enough to 0. This is exactly what is meant by the equation which was to be proved.” (lim sinx/x=1).
That is a correct conclusion, not the squeeze theorem.
Well first of all, $\displaystyle \begin{align*} \cos{(x)} \end{align*}$ does NOT go to 1 when $\displaystyle \begin{align*} x \to \infty \end{align*}$ (in fact, that limit is undefined), but I suspect that is a typo and should be 0.
And now I understand that the reason you are having so much trouble with the squeeze theorem in this case. It's because your definition is weak.
The FULL definition of the squeeze theorem is this:
The squeeze theorem is formally stated as follows.
Let I be an interval having the point a as a limit point. Let f, g, and h be functions defined on I, except possibly at a itself. Suppose that for every x in I not equal to a, we have:
$\displaystyle g(x) \leq f(x) \leq h(x) $
and also suppose that:
$ \displaystyle \lim_{x \to a} g(x) = \lim_{x \to a} h(x) = L$
Then $\displaystyle \lim_{x \to a} f(x) = L$.
The functions g and h are said to be lower and upper bounds (respectively) of f.
Here a is not required to lie in the interior of I. Indeed, if a is an endpoint of I, then the above limits are left- or right-hand limits.
That's not what it's saying at all. Is English your first language?
It is saying that as long as the inequality holds for all x in the domain specified EXCEPT POSSIBLY AT THAT POINT then it is enough to prove the limit.
I am not saying anything at all about the value of sin(x)/x at x = 0. You are correct, it is undefined there. But the function DOES NOT NEED TO BE DEFINED THERE.
And surely if g(x) < f(x) < h(x) at all points except x = a, then $\displaystyle \begin{align*} g(x) \leq f(x) \leq h(x) \end{align*}$ for all points there too.
Do you understand that "less than" implies "less than or equal to"?