He should've taken the absolute value of everything before dividing by sin(x) because if you do, you can prove that it works for a triangle from any quadrant, and in the end he can remove the absolute value of all of them because he will end up with 3 even functions.
He divided by sinx because we know that finally, we are going to apply a limit to sinx where x approaches 0 and therefore sinx cant actually be 0 in this case... what I will agree with is that the method is a convenient fixture, but as convenient as it may seem, it is not wrong mathematically. Also, this is a really good method to explain the property, but of course, there are many more methods to do so.
You can actually use a basic triplet to solve the area of the sector of the circle. Hopefully, you'll agree that the bigger the angle is, the greater the area of the sector we get. We know that angle 2π is propotional to the area of the circle, which is πR² (in this case R=1 which implies that the area is just π). We also know that the angle x is proportional to the area Asub2. Thus, we receive (x/2π)=(Asub2)/π and so A=x/2.
@mephatboi lim x->0 means that x is approaching 0 from both the + and - side. A limit does not exist unless it exists from both sides. Feel free to message me if you have any questions
@mephatboi Great question! If you go to khan's video and watch starting at 12:50 he explains why you don't need the absolute values. Here is the reasoning: If you look at the final inequality over -pi/2 < x < pi/2 then sinx/x and cosx are always positive. That means the inequality is valid for all non-zero x in that interval. This is true even if you don't add the absolute value to begin with. I think he put them in to help guide the viewer, but they were not necessary.
Very good explanation. If you want to proof derivative of sin, this limit also pop's up (and yo can't use L'Hopital then because you are in the way to proof the derivative).
I wonder the same thing. My guess is that the person worked backward. Given that the limit can obtained by numerical means and that the limit as cos(x) approaching 0 is also known as being 1, constructing a geometric argument is obtainable.
Thank you so beyond much. Appreciate it very much and your video is so clean and well put together and I like how chill you are, loved listening to you as it sounded like you were trying to teach me smth irl. Thanks a lot sir.
To go from step 3 to step 4 it is assumed that sin x is larger than zero. If sin x is smaller than zero (when x approaches zero from the negative side), then the inequality signs "flips". However, the end result will also in that case be that sinx/x is between one and one in the limit, so the result stands. Just a little mathematical rigor :-)
Haha kinda late, but here it is: 1 > 1/2 > 1/4 that is true but the reciprocal is not in order to be true you have to change the signs 4 > 2 > 1 That is the simplest way to explain it (I know it will not work for you but just in case someone else have the same doubt)
Arthur Loginow: Late reply here, for anyone questioning why the direction of the inequalities gets changed when reciprocals are taken. If you accept as an axiom that if a > b, then ka > kb whenever k is nonnegative, you will also accept if we multiply by 1/k when k is nonnegative, so if a > b then a/k > b/k. Now use that result but choose k = ab and let's restrict ourselves to situations where ab is nonnegative (a and b have same sign). We can then say if a > b then a/ab > b/ab. but this simplifies to saying if a > b then 1/b > 1/a. Finally, if you prefer you can say if a and b have same sign and a > b then 1/a < 1/b. This captures the idea that if you take reciprocals of the elements of an inequality, you need to switch the direction of the inequality. This would have to be made more rigorous if you wanted to examine instances where a and b are not same sign, but that's not needed in this setting because the expressions here are representing green, pink, blue areas that are all positive, and cos x and sin x are also positive if we assume we are taking the limit as x approaches 0 from the right. In fact, we would not be able to use this idea if different signs were involved. For example, I cannot say if 3 > -2 then 1/3 < -1/2. So use this concept only when both parts of inequality are same sign.
Great explanation. Not sure if anyone already commented, but you didn't explain why the bigger triangle's height is tanx. For those who want to know, one way is to see it as similar triangles, in which sides have to be proportional to each other, and so, tanx/1 = sinx/cosx
Well if they are greater they are certainly greater or equal, so the inequality is still correct. But in the limit we need the equal signs since all the terms become equal.
although it is obvious that measures of those areas are different each other why do we add equality operation between them? İ think they never be equal.Am i wrong?
When we finally take the limit everything does become equal! Before the limit you are right, the areas are strictly greater but since we know we are going to take the limit with throw in the equal sign.
That is squish theorem (sandwich theorem ) if three functions follow the inequality like f(x)≤g(x)≤h(x) Then if lim (x'n a) f(x)=l and lim (x'n a) h(x)=l, therefore lim (x'n a) g(x) should equal l
mohamed, you need to recall the formula to find the area of a sector of a circle... Area of a sector = (1/2)(square of the radius)(central angle) ... in this case, Area = (1/2)(1)(x)... I hope this helps...
@rootmath thanks, i will! so for this, i guess im asking was it unnecessary to take the absolute values of the 3 functions, as salman khan did in his proof (watch?v=Ve99biD1KtA&)?
Man but for 0/0 I can use l hospitals rule And it will result in 1 And sinx is an continuous fuction And Identity function of x is also continuous Therefore sinx/x should also be continuous And at zero it is indeterminate form right So by using limits can't I say it is 1 Ps:) thank u bro ur reply is highly appreciated U r awesome Plz enlighten me if am wrong
@@sdadeveloper5223 you can use l'hopitals rule since you know the derivative of sin(x). But when you try to prove that derivative of sin(x) is cos(x) this exact limit comes up. So without solving this limit we wouldn't know the derivative of sin(x) to begin with.
The equal sign is necessary because if you have 1 < 1 < 1, that wouldn’t be a true inequality. But note that as it concerns the actual functions, it might be true that they are never equal. But the limits are equal. The squeeze theorem doesn’t say that at some point, the functions are equal-sin(x)/x isn’t even continuous at x = 0. It only says their limits are equal.
can you explain how "taking the reciprocal of a whole inequality" is an algebraically-correct operation? im working on a limit proof worksheet that my calc teacher gave us, and it says to convert 1 < x/sinx < 1/cosx to cosx < sinx/x < 1 in two steps is there any way to do it without simply "taking the reciprocal"?
We eventually want to get sin(x)/x in the middle of the inequality. We already have an x at step 3, so we divide by sin(x) to get x/sin(x) and then we flip the fractions (take reciprocals) to get what we wanted sin(x)/x
This demonstration is incorrect. In asserting that the limit (when x -> 0) of cos x is equal to 1, he is using as an argument a result that comes out just from the proof of this theorem, and this is not valid in mathematical logic. That is, the theorem is being demonstrated using the same theorem as an argument.
cos x is a continuous function and cos(0) =1 therefore the limit as x -> 0 of cos(x) = 1 is perfectly valid and does not need this theorem as its justification. The demonstration is correct I just want to make that clear so your comment does not confuse people.
![Proofs: Lim sinx/x =1 and lim [cosx -1] /x =0 as x goes to zero from geometry](http://i.ytimg.com/vi/W0jiodfVGx0/mqdefault.jpg)
I spent so much time trying to understand this on my own when I could have just looked for this video. Thank you, sir, for the post.
You're welcome, glad it helped!
How can you /sinx and then proof, that x-> (cause sin (0)= 0) ->
He should've taken the absolute value of everything before dividing by sin(x) because if you do, you can prove that it works for a triangle from any quadrant, and in the end he can remove the absolute value of all of them because he will end up with 3 even functions.
VeryLazyAngel don't all the negatives just cancel anyway? Even if we consider the 4th quadrant I mean.
He divided by sinx because we know that finally, we are going to apply a limit to sinx where x approaches 0 and therefore sinx cant actually be 0 in this case... what I will agree with is that the method is a convenient fixture, but as convenient as it may seem, it is not wrong mathematically. Also, this is a really good method to explain the property, but of course, there are many more methods to do so.
This was the best explanation I could find for this proof! Thank you for explaining it in a simple manner :)
Heros don't wear capes , they use paint on the PC to explain an equation I was struggling with for weeks 😂❤ thanks man
You can actually use a basic triplet to solve the area of the sector of the circle. Hopefully, you'll agree that the bigger the angle is, the greater the area of the sector we get. We know that angle 2π is propotional to the area of the circle, which is πR² (in this case R=1 which implies that the area is just π). We also know that the angle x is proportional to the area Asub2. Thus, we receive (x/2π)=(Asub2)/π and so A=x/2.
@mephatboi lim x->0 means that x is approaching 0 from both the + and - side. A limit does not exist unless it exists from both sides. Feel free to message me if you have any questions
@snmTube Good observation! I suppose we could break it into two cases, but as you point out the case where sin x < 0 works out the same in the end.
@mephatboi Great question! If you go to khan's video and watch starting at 12:50 he explains why you don't need the absolute values. Here is the reasoning:
If you look at the final inequality over -pi/2 < x < pi/2 then sinx/x and cosx are always positive. That means the inequality is valid for all non-zero x in that interval. This is true even if you don't add the absolute value to begin with. I think he put them in to help guide the viewer, but they were not necessary.
well done and probably the best.
He does not derive the Sector area Area of circle = pi R^2 so Sector area = X/2 pi . pi R^2 = X/2.1^2 = X/2
Great explanation, very intuitive and easy to follow. Thanks!
Very good explanation. If you want to proof derivative of sin, this limit also pop's up (and yo can't use L'Hopital then because you are in the way to proof the derivative).
i want to know how the proof was introduced by that person who basically proved this .......... i want to know how could such a great idea come please
I wonder the same thing. My guess is that the person worked backward. Given that the limit can obtained by numerical means and that the limit as cos(x) approaching 0 is also known as being 1, constructing a geometric argument is obtainable.
Thank you so beyond much. Appreciate it very much and your video is so clean and well put together and I like how chill you are, loved listening to you as it sounded like you were trying to teach me smth irl. Thanks a lot sir.
Maybe I should change my channel to ChillMath 😎
@@rootmath definitely
I loved it! Great proof, i understand it very well, Thank you so much sir, you are a great teacher of mathematics!
To go from step 3 to step 4 it is assumed that sin x is larger than zero. If sin x is smaller than zero (when x approaches zero from the negative side), then the inequality signs "flips". However, the end result will also in that case be that sinx/x is between one and one in the limit, so the result stands. Just a little mathematical rigor :-)
😋😋
Why there is less than or " equal to" sign ? Why equal to ??????
Very nice teaching. All doubts have been cleared. Thanks Sir
You just memorized it without understanding why it was done.
Great explaination.. Brilliant .. 👍
@dweepcan the limit of 1 as x approaches zero is 1. unless you can find someplace to shove that 0 to make the limit 0...
This makes more sense than the numeric method
you mention that you going to explain the equal sign in the inequality , Thanks
Haha kinda late, but here it is:
1 > 1/2 > 1/4 that is true but the reciprocal is not in order to be true you have to change the signs
4 > 2 > 1 That is the simplest way to explain it
(I know it will not work for you but just in case someone else have the same doubt)
Arthur Loginow: Late reply here, for anyone questioning why the direction of the inequalities gets changed when reciprocals are taken. If you accept as an axiom that if a > b, then ka > kb whenever k is nonnegative, you will also accept if we multiply by 1/k when k is nonnegative, so if a > b then a/k > b/k. Now use that result but choose k = ab and let's restrict ourselves to situations where ab is nonnegative (a and b have same sign). We can then say if a > b then a/ab > b/ab. but this simplifies to saying if a > b then 1/b > 1/a. Finally, if you prefer you can say if a and b have same sign and a > b then 1/a < 1/b. This captures the idea that if you take reciprocals of the elements of an inequality, you need to switch the direction of the inequality.
This would have to be made more rigorous if you wanted to examine instances where a and b are not same sign, but that's not needed in this setting because the expressions here are representing green, pink, blue areas that are all positive, and cos x and sin x are also positive if we assume we are taking the limit as x approaches 0 from the right.
In fact, we would not be able to use this idea if different signs were involved. For example, I cannot say if 3 > -2 then 1/3 < -1/2. So use this concept only when both parts of inequality are same sign.
Great explanation. Not sure if anyone already commented, but you didn't explain why the bigger triangle's height is tanx.
For those who want to know, one way is to see it as similar triangles, in which sides have to be proportional to each other, and so, tanx/1 = sinx/cosx
At 5:10 you say, "let's divide everything by sin(x)" but you do not address the fact that sin(x) could be
It didn't rlly matter anyway but yes that is something to consider
I think you have made a mistake when calculating the area of the blue triangle when you put the opposite side is equal to tan(x),
nah.
Can’t be, simply use inverse trig functions to solve for the side length and you get tanx
@ rootmath - The title is written wrong. You are missing grouping symbols. It needs to be written at least as *Proof of sin(x)/x.*
Why did you put equal to the inequality?The triangles are one bigger than the other not equal
Well if they are greater they are certainly greater or equal, so the inequality is still correct. But in the limit we need the equal signs since all the terms become equal.
@@rootmath Ohh i get it.Nice proof!
Is it ok to not to take equal to sign just greater than sign?
Why did you put the less than or equal to sign while we know that it is less than?
I don't understand how you have calculated sector's area.
I mean,normally, don't we calculate (x/360).(r.r.pi)?
u didnt explain the outer limits approaching 0 from the - side
although it is obvious that measures of those areas are different each other why do we add equality operation between them? İ think they never be equal.Am i wrong?
When we finally take the limit everything does become equal! Before the limit you are right, the areas are strictly greater but since we know we are going to take the limit with throw in the equal sign.
@@rootmath it is all about sandwich theorem i guess thank you for your time ;)
amazing proof
And addition to these things, why are we using greater than or equel to less than or equel to? Don't we use greater than or less than directly?
I am looking for its answer too, lmk if youve found it
can i please know why we put the equal sign in the inequality?
That is squish theorem (sandwich theorem ) if three functions follow the inequality like f(x)≤g(x)≤h(x)
Then if lim (x'n a) f(x)=l and lim (x'n a) h(x)=l, therefore lim (x'n a) g(x) should equal l
Wow. That is great!!
thank you teacher but i don t understand the red area
mohamed, you need to recall the formula to find the area of a sector of a circle... Area of a sector = (1/2)(square of the radius)(central angle) ... in this case,
Area = (1/2)(1)(x)... I hope this helps...
Joel Capiral Thank you for this hlep
central angle have to be in radians
why there's an equal sign under > / < signs
but not Green Area < Pink Area < Blue Area?
You are awesome bro 😎 Keep it up
@rootmath thanks, i will!
so for this, i guess im asking was it unnecessary to take the absolute values of the 3 functions, as salman khan did in his proof (watch?v=Ve99biD1KtA&)?
Why you put equal to the areas at step 1? 1:39
Don't you think that area of RHS of green triangle missed out?
شكرآ
this is a sexy proof
REALLY AWESOME VID
why is the limit of a constant 1
the limit of any constant is 0
OK man what will be sinx/x when it is exactly zero
Please help me out bruuh
When x is zero then you have sin(0)/0=0/0 which is undefined.
Man but for 0/0 I can use l hospitals rule
And it will result in 1
And sinx is an continuous fuction
And Identity function of x is also continuous
Therefore sinx/x should also be continuous
And at zero it is indeterminate form right
So by using limits can't I say it is 1
Ps:) thank u bro ur reply is highly appreciated
U r awesome
Plz enlighten me if am wrong
@@sdadeveloper5223 you can use l'hopitals rule since you know the derivative of sin(x).
But when you try to prove that derivative of sin(x) is cos(x) this exact limit comes up. So without solving this limit we wouldn't know the derivative of sin(x) to begin with.
What if u hadnt written less or equals to sign instead of only less than sign
The equal sign is necessary because if you have 1 < 1 < 1, that wouldn’t be a true inequality. But note that as it concerns the actual functions, it might be true that they are never equal. But the limits are equal. The squeeze theorem doesn’t say that at some point, the functions are equal-sin(x)/x isn’t even continuous at x = 0. It only says their limits are equal.
I just don't understand how the height of the blue triangle is tanx.
It's not obvious which I why I made this follow up video explaining it: czcams.com/video/Vve4qCCZPWA/video.html
Cool proof, thanks :)
Is it sandwiches theorem?
thumbsup! Its very clear!!!
thx a lot
can you explain how "taking the reciprocal of a whole inequality" is an algebraically-correct operation? im working on a limit proof worksheet that my calc teacher gave us, and it says to convert
1 < x/sinx < 1/cosx
to
cosx < sinx/x < 1
in two steps
is there any way to do it without simply "taking the reciprocal"?
Taking the reciprocal is basically raising both sides to the -1st power
Thank you!
THANK YOU!!!! :D
quite good
perfect !
Trinogometry
Trinidad and Tobago
HOW IS THE NAME OF THIS PROGRAM ? PLEASE.
@platazar MS Paint?
Mucho bueno
ya know, in the third triangle, why didn't you use sin x instead of tan x for the height of the triangle? you get the answer faster.
how?
I'm te ngarti
Step 3.!!
why.? Divided By sinx
We eventually want to get sin(x)/x in the middle of the inequality. We already have an x at step 3, so we divide by sin(x) to get x/sin(x) and then we flip the fractions (take reciprocals) to get what we wanted sin(x)/x
rootmath Ok.!!Thank
This demonstration is incorrect. In asserting that the limit (when x -> 0) of cos x is equal to 1, he is using as an argument a result that comes out just from the proof of this theorem, and this is not valid in mathematical logic. That is, the theorem is being demonstrated using the same theorem as an argument.
Pedro Vicente RM , Frosty . .. then show us the logically correct way to the proof ! By showing his errors visually . ...
cos x is a continuous function and cos(0) =1 therefore the limit as x -> 0 of cos(x) = 1 is perfectly valid and does not need this theorem as its justification. The demonstration is correct I just want to make that clear so your comment does not confuse people.
The discussion is interesting. Answer me this question: How do you show me, mathematically, that the cosine function is continuous at x = 0?
@@aulavirtualpedrovicenteuse epsilon delta
Please don't make the proof complicated
Sir but the area of a sector is
(x)πr²/360
He's using radians. You are using degrees
Surely not helpful.....
nhổ