Much better than the one in 2008. My only suggestion would be to deal with the absolute values on the front end by proving that lim(x -> 0-)(sin x)/x = lim(x -> 0+)(sin x)/x by noting that f(x)=sin x and g(x)=x are odd functions.
Great lesson! My professor actually requires us to be able to prove that also the cosine function approaches 1 as theta approaches 0, with the definition of limits. Nevertheless, I liked this lesson more than what we've been doing in university. It often feels like lecturers are quite hard to follow in their reasoning. This made it all a tad less complicated!
Where did the right half of the pink triangle come from? The triangle that has sin theta as its left vertical leg. He just draws it out of nowhere 2:08
came across this in Hugh Neil's Calculus book, and i see the author forget to take the last step of taking the reciprocals and inversing the direction of inequality operators... been pondering how to do it otherwise... and the end is the application of sandwich principle
That is a beautiful mathematical truth: as x (in radians) approaches zero, sine x is smaller than but becomes indistinguishable from x. At even at a value as large as 1*10^-5, x and sin x are indistinguishable on my 10-digit calculator. As x gets still smaller and still closer to zero, sin x gets even closer to x. Therefore, the limit of sin x / x^0.5 is x^0.5. (My calculus textbook problem 2.5-supp #25, p166, Howard Anton, Calculus: A New Horizon 6th Ed., incorrectly states on pA73 the answer to problem 2.5-supp #25, what is the limit of sin x / x^0.5 as x approaches zero from the right, is zero, but that's not true, as this proof demonstrates, the limit is not zero but the square root of x.
I'm having a doubt at 3:22 how can you cancel the both pies where the one in the denominator is 360 degrees where the pie in numerator is a constant which equals to 3.14??Someone please clarify my doubt
The area of a circle is given by the formula πr² π was used for the entire circle's area. So if you only want the area of a sector of the circle it is only a portion of the entire area of the circle. To find the ratio of the sector to the entire circle, divide the radian measure of the sector by 2π. That is Θ/2π. Then multiply that by the area of the circle (Θ/2π)πr² = (Θ/2)r². For example, if you only want a semicircle's area, a semicircle has a radian measure of π. That is π/2π = 1/2 of the entire circle. And that is 1/2 of the circle's area or (1/2)πr² = (π/2)r².
Why would you need to prove that? In my opinion it's fairly obvious abs(sin(x)/cos(x)) will always be positive, no matter what the signs of sin(x) and cos(x) are. And in abs(sin(x))/abs(cos(x)) we divide two terms with a positive sign, so the result will also be positive.
I am not sure how one can put the equility sign with the inequility while comparing the three areas. it's pretty much clear the area of circular arc is greater than (and not greater than equal to) area of smaller triangle and area of bigger triangle is greater than the area of circular arc. How in the world they could be equal for any value of theta unless you are directly assuming in the very start itself and once assume so you won't have any way to proceed further because whole geometry will collapse to the origin. Let me know your views.
If you only consider the 1st quarter, you'd only prove what the lim theta ->0 + (i.e. as theta approaches 0 FROM THE RIGHT) is. However, theta may also approach 0 FROM THE left, which means it'd be in the IVth quadrant. By using the absolute values, you can essentially prove that both limit approaching from the left and the limit approaching from the right are equal to 1. saves time :)
Why do we say that the areas are less than or equal to one another? Is is not obvious that the areas are larger than the other so it's more like "less than" not "less than or equal to"? Can someone help? Thanks!
bcs when the theta aproaches zero, everything seems so small and almost equal. For example, let theta is 0.01 degree. Everything in rasio almost equal. Since cos almost 0 degree is almost 1, so sin theta/theta is almost 1 also. What you see is the theha ofc far bigger than 0 let's say 45 degree and it's not approaching zero
Less then or equal to include the case of theta=0 at the moment when you define those areas. But you can exclude theta=0 right away and go with “less than”. The proof will work fine anyway.
Please help me. I am confused. He just prove this only for 1st and 4th Quad... How to prove it for any Quad. Moreover to prove lim(sinx/x), we need to prove it through sine series which is dependent on Maclaurin Series which is dependent on derivative of sin x which is dependent on lim(sinx/x). Then it's a cyclic way. Which is the fundamental theorem and how to prove it without depending on daughter theorems.
I think you just need to prove it in the 1st an 4th because it's lim-0, which means that you want to know how much is sen(0)/0, and sen(0) just appears in the 1st and 4th quad. Sorry for my English I hope that I helped you
the value of sinx approaches the value of x as sinx approaches zero. This means the limit of sinx/x as it approaches 0 will basically be x/x, equalling 1. Sinx approaches x because when sinx approaches 0, x also approaches 0, meaning they become closer and closer to eachothers' values.
If anyone can explain why we put equal sign when it's obvious that the three areas aren't the same and every one much bigger than the other Another thing , can we prove that with any circle rather than unit circle?
Since we didn’t mention the value of theta,it can be any value as long as it’s between -(pi/2) and pi/2 (meaning theta can be any value of the values that are bigger than -(pi/2) and smaller than (pi/2)),and because zero is among these values then currently theta can be zero and when theta is zero the three areas will all be zero since they all equal to sin(theta),theta and tan(theta),therefore the areas are only equal to each other when theta is zero(they are also equal to each other when theta is -(pi/2) and pi/2 but we mentioned earlier that -(pi/2) and pi /2 are not among the values they are just bigger or smaller than that)and it can be done on any circle just put r (which is radius) instead of one and it will give the same result.
When you take the reciprocals you have to invert the signs. Think of an inequality like 1>1/2 when you take the reciprocal and dont change the sign it becomes 1>2 which is obviously incorrect so you have to invert the sign making it 1
I am confused by one thing ...why did we take equality when we were working with equations of areas we can clearly see that the areas are small ...how can they be equal in any case?
I watched this video two years (older version). But after watched multiple times, I still wonder why these three areas are equal, from the graph the blue triangle area should be larger than the red triangle area. Can anyone explain to me?
It *is* equal only when the angle is close to zero. Seems counter-intuitive if you see it through the picture he had drawn. Imagine if you shrink θ to almost nothing, you will see how they have similar areas. If you still think that there might be discrepancies in terms of areas, well, hence the name 'limit'.
@@alkalinewaterman1172 Cause if you go higher than π/r or lower than -π/r, the triangle ceases to be a right-angle triangle, thus screwing up the formulas.
datoubi I suppose it has to do with theta approaching smaller angles the difference in area between the triangles becomes smaller and smaller approaching the same area when theta is 0, but that’s just guessing an explanation
@@braianpita6385 what are the odds of this happening. I was watching this video again right as you responded to my comment :D I think it's really just the limit now as to why it's less or equal to
TheFarmanimalfriend consider the limit as theta approaches zero for both sin theta and tan theta. You’ll see that both approach zero. Graph them or use a graphic calculator, is easy to see. For very small theta the “equality part” in the inequality holds
To use l'Hopital, you must calculate derivatives. And the proof that the deriviative of sine is cosine, requires this limit to be 1. You therefore must proove that separately, as done in this video.
The ending is so perfect.
So this must be equal to one.
And we are done.
Easy when it is so well explained. And it reveals how pure math can be.
I am currently watching this from the older playlist of calculus....he has finally bought a new pen...
He did it like 3 years ago tho
Wow that ending left me feeling very good inside, a happy ending you could say!
Much better than the one in 2008. My only suggestion would be to deal with the absolute values on the front end by proving that
lim(x -> 0-)(sin x)/x = lim(x -> 0+)(sin x)/x
by noting that f(x)=sin x and g(x)=x are odd functions.
That ending is perfect. Talk about going full circle
Great lesson! My professor actually requires us to be able to prove that also the cosine function approaches 1 as theta approaches 0, with the definition of limits. Nevertheless, I liked this lesson more than what we've been doing in university. It often feels like lecturers are quite hard to follow in their reasoning. This made it all a tad less complicated!
That's trivial: the cosine function is continuous at 0. Therefore, by definition, lim theta -> 0 cos theta = cos 0 = 1.
Love the abrupt ending. Gottem.
Wow this is beautiful. I love math!
Where did the right half of the pink triangle come from? The triangle that has sin theta as its left vertical leg. He just draws it out of nowhere 2:08
he's just making a triangle with height sin(θ) and base 1
That entire pink triangle is not right angle triangle ... So (1/2)BH is not applicable !!... Am I right?
@@AfsalPA 1/2(BH) works for all triangles
@@cruinneas5624 Nope. It works only for right triangles. Heron's formula works for all kinds.
@@shigababu8132 can you show me some evidence to why 1/2 x base x height won’t work for any triangle?
came across this in Hugh Neil's Calculus book, and i see the author forget to take the last step of taking the reciprocals and inversing the direction of inequality operators... been pondering how to do it otherwise... and the end is the application of sandwich principle
Why is TH-cam censoring the word function? Whenever Sal says function the audio cuts out? 8:22
Awesome video!! Thanks!
Absolutely wonderful
That is a beautiful mathematical truth: as x (in radians) approaches zero, sine x is smaller than but becomes indistinguishable from x. At even at a value as large as 1*10^-5, x and sin x are indistinguishable on my 10-digit calculator. As x gets still smaller and still closer to zero, sin x gets even closer to x. Therefore, the limit of sin x / x^0.5 is x^0.5. (My calculus textbook problem 2.5-supp #25, p166, Howard Anton, Calculus: A New Horizon 6th Ed., incorrectly states on pA73 the answer to problem 2.5-supp #25, what is the limit of sin x / x^0.5 as x approaches zero from the right, is zero, but that's not true, as this proof demonstrates, the limit is not zero but the square root of x.
Thank you sm!
Thanks
Pretty simple and good
math is amazing thank you
Thank you very much..
I'm having a doubt at 3:22 how can you cancel the both pies where the one in the denominator is 360 degrees where the pie in numerator is a constant which equals to 3.14??Someone please clarify my doubt
Theta is measured in radians. 2pi radians = 360 degrees. pi radians = 180 degrees.
Indeed, pi radians = 3.14 radians. 2pi radians = 6.28 radians.
The area of a circle is given by the formula
πr²
π was used for the entire circle's area. So if you only want the area of a sector of the circle it is only a portion of the entire area of the circle.
To find the ratio of the sector to the entire circle, divide the radian measure of the sector by 2π. That is Θ/2π. Then multiply that by the area of the circle (Θ/2π)πr² = (Θ/2)r².
For example, if you only want a semicircle's area, a semicircle has a radian measure of π. That is π/2π = 1/2 of the entire circle. And that is 1/2 of the circle's area or (1/2)πr² = (π/2)r².
thanks dude!!
Beautiful! Bravo :)
actually at 5:25 you would have to proove that abs(sin/cos) = abs(sin)/abs(cos)
Why would you need to prove that? In my opinion it's fairly obvious
abs(sin(x)/cos(x)) will always be positive, no matter what the signs of sin(x) and cos(x) are. And in abs(sin(x))/abs(cos(x)) we divide two terms with a positive sign, so the result will also be positive.
well, you just made a proof :)
Intelligence is important, but nice abs are *importanter*
Absolute value properties clearly state |x/y|= |x|/|y|
or you could just prove it for the first quadrant and call it a day
Thanks so muchhhh❤
And we are done...
I am not sure how one can put the equility sign with the inequility while comparing the three areas. it's pretty much clear the area of circular arc is greater than (and not greater than equal to) area of smaller triangle and area of bigger triangle is greater than the area of circular arc. How in the world they could be equal for any value of theta unless you are directly assuming in the very start itself and once assume so you won't have any way to proceed further because whole geometry will collapse to the origin.
Let me know your views.
theta can be zero, in which case all would be equal
please update your playlist
Why to bother with absolute values? If it's true in the 1st quarter, it's enough. The forth quarter is just the mirror image.
We're looking for what it is "at" zero. and zero is inbetween the two quadrants.
If you only consider the 1st quarter, you'd only prove what the lim theta ->0 + (i.e. as theta approaches 0 FROM THE RIGHT) is. However, theta may also approach 0 FROM THE left, which means it'd be in the IVth quadrant. By using the absolute values, you can essentially prove that both limit approaching from the left and the limit approaching from the right are equal to 1. saves time :)
Can someone explain why do we need to have the equal sign when comparing areas? It is clear that the areas won't be equal.
how does he make this video? what tools? Thank you!!!
Ok😊
guys, why in the inequality sign he puts less or equal than instead of less than? min 4:24
when theta = 0 all of the areas are 0 and same
In general, divisions are dangerous. Why you can divide by sin(θ) knowing that it can be zero?
Can we prove that the are of the circular sector is bigger than the area of the salmon triangle?
Why do we say that the areas are less than or equal to one another? Is is not obvious that the areas are larger than the other so it's more like "less than" not "less than or equal to"? Can someone help? Thanks!
bcs when the theta aproaches zero, everything seems so small and almost equal. For example, let theta is 0.01 degree. Everything in rasio almost equal. Since cos almost 0 degree is almost 1, so sin theta/theta is almost 1 also. What you see is the theha ofc far bigger than 0 let's say 45 degree and it's not approaching zero
Less then or equal to include the case of theta=0 at the moment when you define those areas. But you can exclude theta=0 right away and go with “less than”. The proof will work fine anyway.
why can't we simply say y = lim ( x->0 ) (sin x)/x ~~ lim (x-->0 ) y = x/x = 1 since for very small x , sinx = x ( also = tan x) ?
Please help me.
I am confused. He just prove this only for 1st and 4th Quad... How to prove it for any Quad.
Moreover to prove lim(sinx/x), we need to prove it through sine series which is dependent on Maclaurin Series which is dependent on derivative of sin x which is dependent on lim(sinx/x). Then it's a cyclic way. Which is the fundamental theorem and how to prove it without depending on daughter theorems.
I think you just need to prove it in the 1st an 4th because it's lim-0, which means that you want to know how much is sen(0)/0, and sen(0) just appears in the 1st and 4th quad. Sorry for my English I hope that I helped you
Is this true for theta is not between -pi/2 and pi/2?
That's irrelevant, since the question here concerns the limit as theta -> 0. Theta can only approach 0 from regions where -pi/2 < theta < pi/2.
Neat proof, but your explanation of tan leaves something to be desired.
At 4:20 how can 3 areas be equal? Clearly second one is bigger than the first and third one is bigger than the second.
Notice when theta = 0, all three areas are 0, so 0
Why is 4:25 the area of the salmon triangle less than OR EQUAL to the area of the wedge? I don't get it how can they be equal can anyone explain?
As theta approaches 0 the value of all three areas should approach to equal value
the value of sinx approaches the value of x as sinx approaches zero. This means the limit of sinx/x as it approaches 0 will basically be x/x, equalling 1. Sinx approaches x because when sinx approaches 0, x also approaches 0, meaning they become closer and closer to eachothers' values.
You can use the definition of derivative of sin(x) at point 0
No. You need lim(x->0)sinx/x to find the derivative of sine. Try to compute derivative of sine by definition.
If anyone can explain why we put equal sign when it's obvious that the three areas aren't the same and every one much bigger than the other
Another thing , can we prove that with any circle rather than unit circle?
because of sandwich or squeeze theorem.
Since we didn’t mention the value of theta,it can be any value as long as it’s between -(pi/2) and pi/2 (meaning theta can be any value of the values that are bigger than -(pi/2) and smaller than (pi/2)),and because zero is among these values then currently theta can be zero and when theta is zero the three areas will all be zero since they all equal to sin(theta),theta and tan(theta),therefore the areas are only equal to each other when theta is zero(they are also equal to each other when theta is -(pi/2) and pi/2 but we mentioned earlier that -(pi/2) and pi /2 are not among the values they are just bigger or smaller than that)and it can be done on any circle just put r (which is radius) instead of one and it will give the same result.
THANK YOUUUUUUUU
I dont understand one thing.
The area of the triangle can be less than the area of radian but when will the areas will be equal?
Plz help.
When Theta = 0
Why did signs flip when sin θ/θ was inversed?
When you take the reciprocals you have to invert the signs. Think of an inequality like 1>1/2 when you take the reciprocal and dont change the sign it becomes 1>2 which is obviously incorrect so you have to invert the sign making it 1
Taking the reciprocal reverses the sign of the inequality. It's a general math rule. You are reversing the inequality essentially.
Wow
is sal doing a barack obama voice imitation at some points in the video
why is absolute value used?
I am confused by one thing ...why did we take equality when we were working with equations of areas we can clearly see that the areas are small ...how can they be equal in any case?
As theta approches zero the values of all three areas should approach to equality..
Where the base =1 come from?
published this pretty recently. Anyways why do you need to change the signs upon using the recipricals
Ty appreciate the explanation
davidiswhat was a day
I watched this video two years (older version). But after watched multiple times, I still wonder why these three areas are equal, from the graph the blue triangle area should be larger than the red triangle area. Can anyone explain to me?
It *is* equal only when the angle is close to zero. Seems counter-intuitive if you see it through the picture he had drawn. Imagine if you shrink θ to almost nothing, you will see how they have similar areas. If you still think that there might be discrepancies in terms of areas, well, hence the name 'limit'.
@@agastyawiraputra2208 Thank You!
@@agastyawiraputra2208 Can you explain me why has he chosen the interval as -π/2
I also came across this question
@@alkalinewaterman1172 Cause if you go higher than π/r or lower than -π/r, the triangle ceases to be a right-angle triangle, thus screwing up the formulas.
Also I dont see why the pink triangle could be equal to the orange slice of the circle.
They are equal if theta = 0.
The triangle will have an area of 0, and the circle slice will also have an area of 0
0=0
Me too , I can't get this step at all
It's doesn't make any sense
Ahmed Azmy
I literally just explained it in the comment above...
oh !! thanks ♥ @attila th fun
why is it less or equal to and not just less than? I cant see how the red triangle could ever be equal to the orange triangle
datoubi I suppose it has to do with theta approaching smaller angles the difference in area between the triangles becomes smaller and smaller approaching the same area when theta is 0, but that’s just guessing an explanation
@@braianpita6385 what are the odds of this happening. I was watching this video again right as you responded to my comment :D I think it's really just the limit now as to why it's less or equal to
@@datoubi glad I could help :)
How can the area be less or equal
Thx
epic
we are done lmao
According to your logic, sin theta/2 = tan theta/2 (area of a triangle). I have a problem believing that.
TheFarmanimalfriend consider the limit as theta approaches zero for both sin theta and tan theta. You’ll see that both approach zero. Graph them or use a graphic calculator, is easy to see. For very small theta the “equality part” in the inequality holds
wrong. according to the logic the LIMITS as theta ->0 of sin theta / 2 = LIMIT as theta ->0 of tan theta /2. And that IS true.
UAO
Why take the reciprocal?
DeRon Burton To get rid of 1/cosθ and make it easier to calculate
because the question was "what's the limit of sin x / x", not "what's the limit of x / sin x".
i don't get it
how am I supposed to do such reasoning in the exam
L'opital gives cos(x)/1 from sin(x)/x and that is equal to one. as x --> 0.
To use l'Hopital, you must calculate derivatives. And the proof that the deriviative of sine is cosine, requires this limit to be 1. You therefore must proove that separately, as done in this video.
some weird math about how 0/0 is 1 interesting.
check out rootmath