The Most Important Limit In Calculus

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Disclaimer: This video is for entertainment purposes only and should not be considered academic. Though all information is provided in good faith, no warranty of any kind, expressed or implied, is made with regards to the accuracy, validity, reliability, consistency, adequacy, or completeness of this information.

#math #brithemathguy #limit

[BriTheMathGuy]

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[yihongzhu]

As an engineer, it’s simple, sin(x)=x, so x/x=1. QED

[nonameAccountable]

At 3:01 you can divide all sides by sin(x) only because it's limit, otherwise it would be division by 0.

[Mutual_Information]

A geometric approach to a famous limit.. very nice

[NintendoGamer]

A similar iequality to apply squeeze theorem to can arise from just analyzing lengths. If we consider a simple right triangle's height inside the circle, the arc length of that sector, and the height of the larger right triangle sticking outside of the circle, it can be seen that sinx ≤ x ≤ tanx (which simplifies to the same thing in the end).

[Pedro-dnsg]

Just to point out, I think L'Hôpital's rule here is a bit problematic. That is because, by the definition of a derivative, a similar limit would appear in d(sin x)/dx, meaning that the concept d(sin x)/dx = cos x builds upon the result of the limit you're trying to solve. The geometric approach with the squeeze theorem in fact seems to be a much better way to solve this.

[natox]

The way I once explained this very intuitive limit (even if it doesn't seems like), is by saying that when you look away to the horizon, you can't see the curvature of the earth, it seems like it's a straight line. That's because we're too small compared to the earth and can't see far enought. So in this case, what we see is sin (x), x being the distance we see, and then x = sin (x) for a very small distance compared to the whole circumference of the earth (x approches 0).

[stevewhitt]

I have studied calculus, but this is the very first time that I have seen this geometric analysis. A++

[mathisnotforthefaintofheart]

The geometric approach is King. All the other "proofs" involving calculus is mere circular reasoning. My question is: Is there also a DIRECT geometric proof for the limit (1-cosx)/x = 0 without using sinx/x or tanx/x ?

[ASHOKKUMAR-buey]

Thanks bro you helped me out to get it!! Can't explain how much you helped me through words 🙂

[thetruecobra]

You can avoid any circular reasoning by defining sin(x) in terms of complex exponentials. Then its derivative will be trivial and l'hopital & other methods requiring the derivative of sin can be used.

[breakoff]

I tried making a video similar to this. How did you construct the triangles and unit circle and make everything link up (software, etc). Love your videos BTW!

[RCSmiths]

Such an interesting way to assess a Limit! Thank you for these very intriguing and engaging videos.

[adityashimoga]

L'Hôpital's rule can also be used

[terryendicott]

Most, if not all, calculus books prove that the derivative of sine is cosine using this limit. So if one were to use either L'Hopital's Rule or the Maclaurin series to get this limit you would first have to get the derivative of sine some other way. Unless, of course, you like circular reasoning - which is a capital offense in mathematics.

[sukhps]

this is a 5 marks question for pre university in India
this exact proof is what i wrote in exam

[AcryllixGD]

Cant you use small angle approximations to prove this as well? As sin x ≈ x, and x is very small so the uncertainty is tending to 0, x/x (where x is not 0) = 1. I could be wrong about this proof, as im a new a level student, but good video nonetheless!

[sutirthjha]

I learned this myself from a graph. Both graphs start from 0 and initially derivative of sin theta is cos 0 = 1. So both graphs overlap fo a short while.

[consciouseco]

If it's close enough it's enough.

[glumbortango]

Month afterward, but if you represent x with the arc length of the circle, it's visually obvious that the two converge to each other.