Understanding Limits: Solving Homework Equations without Quotation Marks

[Karol]

Homework Statement

Homework Equations

The Attempt at a Solution

But i didn't yet learn limits methods. i am not asked here to find the limit.
I only know the function isn't continuous at 0, and this can be fixed here because of graph.
Maybe the book gave this on the base of the drawing.
But later i am asked to prove it

 

[Mark44]

Homework Statement

View attachment 206846

The problem statement is quite vague.
$\lim \frac{\sin(\theta)}{\theta}$ doesn't indicate what $\theta$ is doing.
Based on the limit value, the above should probably be stated as $\lim_{\theta \to 0} \frac{\sin(\theta)}{\theta} = 1$

Homework Equations

View attachment 206845

The Attempt at a Solution

But i didn't yet learn limits methods. i am not asked here to find the limit.
I only know the function isn't continuous at 0, and this can be fixed here because of graph.
Maybe the book gave this on the base of the drawing.
But later i am asked to prove it

What are you attempting to show in the graph? You have what appears to be parabolic in shape, opening downward, and containing the points $(-\pi/2, 2/\pi), (0, 1),$ and $(\pi/2, 2/\pi)$, with no indication of the formula used to generate the graph. It's possible to do a geometric "proof" using the unit circle, with a small angle $\theta$.

 

[Karol]

The graph is $~\lim \frac{\sin\theta}{\theta}$
The book doesn't want graphic proof.
I don't know on which assumptions it's based. the chapter is the second in the book.

 

[Ray Vickson]

The graph is $~\lim \frac{\sin\theta}{\theta}$
The book doesn't want graphic proof.
I don't know on which assumptions it's based. the chapter is the second in the book.

Lots of proofs are available on-line. They all depend in some way on the fact that for an arc of radian angle $\theta$ in a circle of radius $r$ the length of the arc is $L = r \theta$. Some of them are a bit intuitive, depending on believing that the length of a small arc and the length of the subtended line segment have a ratio that goes to 1 as the angle goes to zero. Others are a bit more rigorous, in that they obtain both upper and lower bounds on the ratio $\sin(\theta)/\theta$ and show that both bounds have limit 1 as $\theta \to 0$. See, eg.,
https://math.stackexchange.com/questions/75130/how-to-prove-that-lim-limits-x-to0-frac-sin-xx-1
for a proof of that type.

 

[Mark44]

The graph is $~\lim \frac{\sin\theta}{\theta}$

No, that doesn't make any sense. Possibly the graph is $y = \frac{\sin(\theta)}{\theta}$, but a limit represents a single number, provided that the limit exists.

The book doesn't want graphic proof.
I don't know on which assumptions it's based. the chapter is the second in the book.

I'm not talking about a graphic proof. I'm suggesting that you use the geometry of the unit circle.

Has your book presented proofs of other limits, such as $\lim_{x \to 0}\frac{\cos(x) - 1}{x}$?

 

[Karol]

sin(x) and x are continuous functions and the quotient of continuous functions is continuous. that's why there's a limit.

$$\lim_{\theta\rightarrow 0}\frac{\sin\theta}{\theta}=\lim\frac{h}{\breve{AB}}=\lim\frac{\sqrt{1-\cos^2\theta}}{\theta}$$

 

[Mark44]

sin(x) and x are continuous functions and the quotient of continuous functions is continuous.

$\frac{f(x)}{g(x)}$ is continuous on any interval on which $g(x) \ne 0$. In your problem, $\frac{\sin(x)} x$ isn't defined at x = 0, let alone being continuous at that point.

that's why there's a limit.

A limit can exist independent of whether the expression involved is continuous or even defined at the point in question.

 

[Karol]

If we define f(0)=0, with the frase about $~\frac{f(x)}{g(x)}$, are these the basis for the limit's existence?

 

[Mark44]

If we define f(0)=0, with the frase about $~\frac{f(x)}{g(x)}$, are these the basis for the limit's existence?

No. In the quotient $\frac{\sin(x)}x$, sin(0) is already 0, so there's nothing to gain by defining sin(0) = 0. The problem is that both numerator and denominator are zero when x = 0. As your drawing in post #6 suggests, $\lim_{\theta \to 0} \frac{\sin(\theta)} \theta$ exists, with the limit being 1, but this has very little to do with $\frac{\sin(\theta)}\theta$ being continuous at $\theta = 0$.

 

[Karol]

So how do i know if a function has a limit, without calculating it at first?
I am asked on what basis that limit exists, and they may also mean on what basis it equals 1

 

[Mark44]

So how do i know if a function has a limit, without calculating it at first?

If you can show that the function is bounded above and below by two expressions, and both expressions have the same limiting value, then the limit of the function you're looking at must be the same. This is what @Ray Vickson was saying in post #4. Also take a look at the link he provided.

I am asked on what basis that limit exists, and they may also mean on what basis it equals 1

 

[Karol]

Lots of proofs are available on-line. They all depend in some way on the fact that for an arc of radian angle θθ\theta in a circle of radius rrr the length of the arc is L=rθL=rθL = r \theta

Where can i find such a proof? can you give me a link, or what to write at the search?

 

[Mark44]

Where can i find such a proof? can you give me a link, or what to write at the search?

From post #4

Lots of proofs are available on-line.
<snip>
See, eg., https://math.stackexchange.com/questions/75130/how-to-prove-that-lim-limits-x-to0-frac-sin-xx-1
for a proof of that type.

From post #11

This is what @Ray Vickson was saying in post #4. Also take a look at the link he provided.

Please take the time to read what members have already posted...

 

[Ray Vickson]

Where can i find such a proof? can you give me a link, or what to write at the search?

I cannot believe you are asking that; it is all right there in the post you quoted from!

 

[Karol]

Sorry i was misunderstood. i read every post.
I meant the proof involving $~L = r \theta~$, not the limits method.
I looked at the link Ray gave me, it's good but it's with the limits.

 

[Mark44]

I meant the proof involving $~L = r \theta~$, not the limits method.

That's how arc length is defined. For a full circle of radius 1, $L = 1 \cdot 2\pi$, its circumference.

I looked at the link Ray gave me, it's good but it's with the limits.

What's your question here?

 

[Karol]

Ray said there are many proofs involving the method $~L=r\alpha~$ on the net, how do i find them?

 

[Mark44]

Ray said there are many proofs involving the method $~L=r\alpha~$ on the net, how do i find them?

He said there are many proofs that $\lim_{x \to 0}\frac{\sin(x)}x = 1$. He also said that they all depend on the fact that $L = r\theta$, which is a fact by definition.

 

[Ray Vickson]

Sorry i was misunderstood. i read every post.
I meant the proof involving $~L = r \theta~$, not the limits method.
I looked at the link Ray gave me, it's good but it's with the limits.

I don't understand your objection. Your question was: what is the limit and how do you prove it? The link answered all of that, exactly as you requested.

You can find other treatments just by trying out some key words in Google. I found the articles by entering "limit of sinx/x", but others might work too. Just experiment!

However, I regard many of these arguments as "proofs" rather than proofs; that is, they leave a lot of holes that are not easy to fill in rigorously. That is why in my original response I mentioned "intuitive arguments", which rely on the belief that the length of an arc and the length of a subtended segment have ratio 1 in the limit of small angle. Such arguments are OK in an introductory calculus course, but are unsatisfactory in a more rigorous treatment (unless some other relevant properties of arc-lengths have been proved first---not likely in an intro calculus course).

I think that area-based arguments are more sound, because if one region is entirely contained between two others, its area lies between the inner and outer areas. For arc-lengths, that type of property does not apply directly: if a curved arc lies between an inner polygonal and outer polygonal path, the arc is not "part" of either, and it is not 100% clear that its length lies between the inner and outer lengths. I remember seeing an entire, rather rigorous graduate-level book devoted to such matters as the foundations of arc-lengths, surface areas and the like. Apparently, a number of eminent early 20th century mathematicians were bothered by this issue, and wrote research papers on the topic. Unfortunately, I only saw the book in a library about 30-40 years ago, and no longer remember its title or author.

 

[Karol]

I thank you very much Ray and Mark, and i am sorry i was misunderstood, i value this site and your help very much.
I still don't understand the first part of the question: "under what assumption is true that $~\lim_{x \to 0}\frac{\sin(x)}x = 1~$. the continuity isn't relevant here so what could they mean? the book didn't cover very much until this point.
Oh, now i see your post Ray and i will read it

 

[Mark44]

I still don't understand the first part of the question: "under what assumption is true that $~\lim_{x \to 0}\frac{\sin(x)}x = 1~$. the continuity isn't relevant here so what could they mean?

The question is pretty vague, IMO. Possibly the assumption this author is looking for is that x is a real number -- the radian measure of an angle rather than its measure in degrees.
If you use a calculator with values of x getting closer to 0, such as .1, .01, .001, and so on, you will get different values for $\frac{\sin(x)} x$ when the calculator is in degree mode as opposed to radian mode.

 

[Karol]

Thank you again Mark44