Proof: Limit of sinx/x as x approaches 0 with Squeeze Theorem | Calculus 1
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- Опубліковано 12 чер 2023
- 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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![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)
Get practice using the sinx/x limit here! ua-cam.com/video/iK0FSLZXHBQ/v-deo.html
Calculus 1 Course: ua-cam.com/play/PLztBpqftvzxWVDpl8oaz_Co6CW50KtGJy.html
Calculus 1 Exercises playlist: ua-cam.com/play/PLztBpqftvzxUEqGGgvL3EuIQUNcAdmVhx.html
What software are you using to write and annotate the explanation?
Best explanation on the internet of this theorem. Bravo, sir!😊
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!
@@WrathofMath Picky is good in math! 😁
@@punditgi or sunao pandit ji tum yaha kya kar rahe 😂 btw me too bramhan (purohit ❤)
bro ur so energetic ty
The best explanation so far.
Thank you!
nice explanation, I saw this in my calc textbook but this explains the steps very well
Glad it was helpful!
thanks a lot
I use hopital for the last problem. What are alternative approaches that we can use?
Great exercise, thank you
My pleasure - thanks for watching!
Your videos are excellent!
Thank you!
Great explanation.
Thank you!
What a nice explanation
Thank you!
Can you use the squeeze theorem to prove that the limit as x approaches 0 of (cos(x) - 1)/x is 0?
I loved this video
Thank you!
THE GREATEST VIDEO ON THE INTERNET!!!!!
THANK YOU SO MUCH SIR THIS WAS EXTREMELY HELPFUL
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.* 😅
I finally understand this concept after watching this clear explanation ! Thank you.
Glad to help - thanks for watching!
Great explanation as always...you have knowledge and talent to deliver informatiin...Respect from lebanon
Thank you!
How do you prove that the tan area is larger than the sector area? Since the sector is curved?
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
Very nice, clear explanation, with simple, clear diagrams. Well done.
Many thanks!
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 ?
I would say the fact that sinx is similar to x near 0 is proven by this limit, certainly not the other way around!
@@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.
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.)
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.
Is there any proof for the order of areas?
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!
@@PapaBavarianthanks man
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
wow
Thanks for watching!
Making use of area to derive the inequality is circular reasoning.
Pun intended?
@@zat5176 No, think of how area is being derived and see why it is circular.
@@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
@@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.
@@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