Proofs: Lim sinx/x =1 and lim [cosx -1] /x =0 as x goes to zero from geometry

[@SAIROSHAN-li4bx]

I dont understand why you have so many less subscribers like your videos are so interesting and I just stare at the screen, becuz you made learning so interesting and easy. Thank you. I understood it so well

[@ThePowerfulOne07]

Wow never had this explained to me in such great detail. I always was told because of the squeeze theorem…but to see it like it makes TOTAL SENSE!

[@punditgi]

Prime Newtons always proves he is always worth watching! 😊🎉

[@sujaybhattacharya6562]

Extraordinary lecture. This man's name shall be told.

[@sudiptoatutube]

My heart fills with so much joy after watching your video. This one too was a perfect cocktail of Geometry, Calculus and algebra. Love you bro. God bless you.

[@孫西萍]

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[@HackerBoy-uk3td]

i love you sir you don't know how hard was it to get my head around this but your explanation helped me a lot. I'm your new subscriber and looking forward to learn more from you.

[@AnuragAnand999]

thank you for making my life simpler

[@panyachunnanonda6274]

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[@holyshit922]

I would use double angle identity and Pythagorean identity cos(theta)=cos^2(theta/2) - sin^2(theta/2) 1 = cos^2(theta/2) + sin^2(theta/2)

[@PrimeNewtons]

Yep!

[@frreinov]

Humble suggestion: "this radius 1 circle " is like sorcery for those who start learning trigonometry. You (at least me) thought at first "does this only work on a radius 1 circle? What about all other circles??". So I believe it is useful to explain that any circle can be a radius 1 circle and that you just normalize it to 1 for simplicity. I hope I could pass the idea to you.

[@heiu304]

Try not to use L’Hôpital💀💀💀

[@smolblacquecat7148]

IMPOSSIBLE!!!!!

[@nychan2939]

I noticed something in the second proof: lim[(1-cosθ)/θ][(1+cosθ)/θ]=[lim(1-cosθ)/θ][lim(1+cosθ)/θ] It's the identity lim ab=(lim a)(lim b) We must have both lim a and lim b exist. But lim a is not known yet.

[@cameronconely1304]

Even though you don’t know it exists at the time. You can prove it exists after using the epsilon-delta definition I believe. That being said, I’m just a high school student who isn’t the most mathematically inclined. So I might be wrong

[@SimoncitoReinhardt]

Great teaching skills, and good video, congratulations. But I fear I must have missed something. You are trying to find the proof that lim(th to 0) of [cos(th)-1]/th equals 0 but you end up with lim(th to 0) of [1-cos(th)]/th =0. What I am trying to say is, how come you start with cos(th)-1 and then end with 1-cos(th)? I assume that they are equivalent at this point, but I missed how and why. Could you please help me out with this?

[@jarousskyphilippe5831]

凄いわかりやすくて助かりました！

[@Gargee_Paul]

Superrbb teaching 👍🏻

[@SansWordHuang]

Thanks for the video. My question for the second part is, if we have two number number multiplied to be 1, but one of it is zero in the end, then how it can be one when multiplied to another number?

[@PrimeNewtons]

You should not assume one of them is zero. Remember, the limit is about 'as x approaches a point'. It doesn't mean x gets to that point. Good thinking, though.

[@johnnolen8338]

You can also use the identity: cos (x) - 1 = - 2·sin^2(x/2) to prove lim x --> 0 [cos(x) - 1]/x = 0.

[@chenruizhang7878]

Hey! This is really cool but when proving the special limit for sin, shouldn’t you only take the right hand limit because you set theta to be greater than 0? Then you would use the fact that the sin function is odd to prove that the limit from the left also = 1

[@MrBesmir7]

It crazy idea that math could be so erotica, term squeeze is so 🔥

[@jf3518]

So funny. I watched yesterday your video about derivative of sin and the referred video was not available anymore about the cos identity and here is a new one.

[@sebastianMo2]

You savior!

[@tondysek]

Don't learn the formulas, but the way to find them! Greetings from Bavaria

[@vipin-choubey]

9:46 why assume theta is positive what if it for negative? If we consider theta as negative may be it between -1 to +1 at 11:55

[@oreo_smoothie74]

Because theta is equal along all terms. If it was negative in all of the then we would multiplying the entire inequality by - 1 and get back to the case where theta was posetive in the first place.

[@4iden.r]

I don't understand why it is less than or equal to? How would the area of the sector be equal to the area of the sub triangle? And similarly how would the area of the larger triangle be equal the the sector and the sub triangle? Other than that, this video was awesome!

[@MrBesmir7]

It crazy idea that math could be so erotica, term squeeze is so 🔥

[@schwobelabibalbol]

There is a case when they are equal. When theta = 0+2n pi, where n is an integer

[@mathemitnullplan]

awesome, how did you squeez "Limits" to the description title?

[@guritwulan9826]

so great, I completely understand. thanks sir

[@ThePrana3]

Can you do more proofs?!!!!

[@SpinozaHK]

Question: When we divide the equation by sinΘ, how do we determine that sinΘ > 0? Because when sinΘ ≤ 0, the equation will change. Can someone explain what happened in this step?

[@mr.cow-boy6105]

Thanks for this amazing video 😊

[@mathszone8025]

the quote is very important to the peaple. go on quotation

[@aahanshrivastava1605]

The sin theta part was taught beautifully but I didn't get the cos theta part because we had to prove = cos theta -1/theta=0 but here you got 1-COS THETA/theta = 0?

[@vivekranavaya4043]

Their both the same thing the negative sign just goes to zero

[@aahanshrivastava1605]

@@vivekranavaya4043 oh ok thanks 😁

[@robertveith6383]

You are missing grouping symbols around your numerators.

[@faridaorucova9990]

Why area of sector is not (pi*r^2*theta)/360????

[@Aliumair27278]

What a teaching skill ❤

[@fathimahishma5149]

can i please know why you put the equal sign for the inequality?

[@nychan2939]

1. a>b => a≥b 2. a≥b => a>b 1 is legitimate but 2 is NOT. So feel free to use 1 whenever it's helpful.

[@mathacademyforall4161]

when the equal sign is validated in the inequaity please

[@nychan2939]

1. a>b => a≥b 2. a≥b => a>b 1 is legitimate but 2 is NOT. So feel free to use 1 whenever it's helpful.

[@wanwodsciencedepartment8801]

Please sir, I am from Sierra Leone help with differentiation of Determinant

[@ShreeRadder]

Sir can i write this annual exam??

[@shreyaalurshreyaalur6443]

I wrote it and bhoom . I got full marks 😀😀

[@AngeloLaCruz]

Tremendous approach. Greeting From Curaçao An Island Nation in The Caribbean

[@PrimeNewtons]

Greetings!

[@abheofficial_]

Thank you sir ❤

[@josephacri6855]

Love your video super great on your explanations looking forward to more of your videos,thank you for your great work

[@josephacri6855]

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[@techitnow9626]

It all makes sense simplifying the geometric proof. But after all I'll need a sandwich for my lunch time out.😊

[@masoudhabibi700]

Thanks for an other video....mas than Iran

[@No_prejudice_pls]

Thanks

[@SimbehBah]

Love it

[@oooooommmmmmmmm1857]

No area of sector

[@OfficialNND]

why, whenever i got your point you always looking at me😂, as if you said "Eureka, right?"

[@okanepara159]

watermelon :) If I said it in the classroom I would be kicked out :)

[@thexoxob9448]

11:34 here it doesn't make sense if you want to use the squeeze theorem. In fact it doesn't make sense at all since this is what the squeeze theorem implies. Let me show you what I mean. If 1 0 sin x/x

[@robertveith6383]

Write sin(x)/x. You need grouping symbols.

[@thexoxob9448]

@@robertveith6383 the important thing is you get the point

[@MrBesmir7]

It crazy idea that math could be so erotica, term squeeze is so 🔥

[@luisclementeortegasegovia8603]

Watch out professor, because the area of the sector explained would be: (theta/2π) (π.r²) = theta/2.

[@kingbeauregard]

I can always remember how to prove sinx / x, but (1 - cosx) / x always throws me. I don't think I've seen it solved this way before, or maybe I have but I just can't remember. Either way, I like it!

[@robertveith6383]

Put the argument in grouping grouping symbols: sin(x)/x, [1 - cos(x)]/x

[@profenevarez]

well done! 👏🏼

[@PrimeNewtons]

Thank you!! 😁

[@Desalegn-nj8bo]

thank you very much and much ! keep it up!

[@MazinBugshan]

Excellent

[@dimitriskyriakopoulos1984]

Many and tragic Mathematical errors !!! Ask Mathematician .

[@kova-111]

We are dividing by Zero in the second proof. That cant be right.

[@francaisdeuxbaguetteiii7316]

The integrity of a true mathematician. 🎉

[@PrimeNewtons]

I hope I'm consistent. Thank you 😊

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[@pavithras1858]

Me..

[@Rajesh_edizs]

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