Proof: Limit of sinx/x as x approaches 0 with Squeeze Theorem | Calculus 1
We prove the limit of sinx/x as x goes to 0 equals 1 using the squeeze theorem and a geometric argument involving sectors and triangles on the unit circle. #calculus1 #apcalculus
Squeeze Theorem Explained: (coming soon)
Limit of (1-cosx)/x: • Limit of (1-cos(x))/x ...
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Пікірлер: 56
Get practice using the sinx/x limit here! kzread.info/dash/bejne/m39kqLWFiruucrQ.html Calculus 1 Course: kzread.info/head/PLztBpqftvzxWVDpl8oaz_Co6CW50KtGJy Calculus 1 Exercises playlist: kzread.info/head/PLztBpqftvzxUEqGGgvL3EuIQUNcAdmVhx
Best explanation on the internet of this theorem. Bravo, sir!😊
@WrathofMath
Жыл бұрын
Thank you Ezra! I actually feel like I flubbed parts of this, but I'll leave it up for now and see if people generally find it helpful. I get pickier and pickier as the years go on, so maybe it's just me being a freak!
@punditgi
Жыл бұрын
@@WrathofMath Picky is good in math! 😁
@Lnx435
9 ай бұрын
@@punditgi or sunao pandit ji tum yaha kya kar rahe 😂 btw me too bramhan (purohit ❤)
The best explanation so far.
@WrathofMath
5 ай бұрын
Thank you!
Finally a channel that uses the squeeze theorem correctly 😅
@WrathofMath
Ай бұрын
Thanks for watching!
thanks, this is a great explanation.
@WrathofMath
Ай бұрын
Glad it was helpful! Thanks for watching!
nice explanation, I saw this in my calc textbook but this explains the steps very well
@WrathofMath
6 ай бұрын
Glad it was helpful!
I use hopital for the last problem. What are alternative approaches that we can use?
What software are you using to write and annotate the explanation?
How do you prove that the tan area is larger than the sector area? Since the sector is curved?
@lautamn9096
9 ай бұрын
Because tan area is containing sector area and also sector area doesnt fill the tan area. Thats why this theorem only works with theta approaching 0 i.e. very small angles
bro ur so energetic ty
Can you use the squeeze theorem to prove that the limit as x approaches 0 of (cos(x) - 1)/x is 0?
Your videos are excellent!
@WrathofMath
6 ай бұрын
Thank you!
2:23 For mnemonic purposes, I'm going to note that the area formula with *Sin* is the *Smallest,* while the one with *Tan is the Tallest.* 😅
Great explanation.
@WrathofMath
7 ай бұрын
Thank you!
Great exercise, thank you
@WrathofMath
Жыл бұрын
My pleasure - thanks for watching!
thanks a lot
Nicely done! It's been nearly four decades, but I'm pretty sure this is how I learned it from Apostol. (The book, not the man; he'd retired from teaching freshmen the year before.)
@WrathofMath
Жыл бұрын
Thanks Tom! This is always how I have seen it done in textbooks, though I've never had the pleasure of reading Apostol! I've been shopping for his books recently, they're just so expensive.
Very nice, clear explanation, with simple, clear diagrams. Well done.
@WrathofMath
8 ай бұрын
Many thanks!
What a nice explanation
@WrathofMath
Жыл бұрын
Thank you!
THE GREATEST VIDEO ON THE INTERNET!!!!! THANK YOU SO MUCH SIR THIS WAS EXTREMELY HELPFUL
I finally understand this concept after watching this clear explanation ! Thank you.
@WrathofMath
10 ай бұрын
Glad to help - thanks for watching!
I loved this video
@WrathofMath
Жыл бұрын
Thank you!
Hey it's just a thought but we know sinx for very small values of x is similarly equal to x, right. Then the limit would be lim x-> 0 (x/x) . We can cancel out the x and get 1. Can this be a ideal solution though ?
@WrathofMath
4 ай бұрын
I would say the fact that sinx is similar to x near 0 is proven by this limit, certainly not the other way around!
@daringd2147
4 ай бұрын
@@WrathofMath You are right, I searched on it and came to know that it does come from this limit. Didn't know about it, I just kinda looked at the graphs, and some questions that use this approximation.
Great explanation as always...you have knowledge and talent to deliver informatiin...Respect from lebanon
@WrathofMath
Жыл бұрын
Thank you!
limit((1-cos(x))/x,x=0) = limit((cos^2(x/2)+sin^2(x/2)-cos^2(x/2)+sin^2(x/2))/x,x=0) =limit(2sin^2(x/2)/x,x=0)=limit(sin(x/2),x=0)*limit(sin(x/2)/(x/2),x=0)=0*1
Is there any proof for the order of areas?
@PapaBavarian
5 ай бұрын
Good question, and this part of the otherwise excellent proof here is guilty of 'hand waving.' But yes, the best way is to do a proof by contradiction. Just assume that the order of the areas are not as he states and you'll find contradictions which will prove that the area inequalities are valid!
@dark3l192
5 ай бұрын
@@PapaBavarianthanks man
wow
@WrathofMath
5 ай бұрын
Thanks for watching!
Making use of area to derive the inequality is circular reasoning.
@zat5176
10 ай бұрын
Pun intended?
@yemoeaung251
10 ай бұрын
@@zat5176 No, think of how area is being derived and see why it is circular.
@immutabledestiny6377
9 ай бұрын
@@yemoeaung251you need to explain why that is, because there is nothing occurring which is circular reasoning, and the justification for the inequalities arises because of basic geometry and geometric arguments
@yemoeaung251
8 ай бұрын
@@immutabledestiny6377 The derivation of area involved that limit itself. A lot of the textbooks out that are doing this proofs which is not rigorous at all.
@gamesandthoughts2388
8 ай бұрын
@@yemoeaung251People just don't care. What's in the video is more of an illustration, rather than a rigorous proof. Although, some of the book authors sometimes note that this illustration is not a proof with the same reasoning that u said