Calculus: Arnold's limit

The Russian mathematician Vladimir Arnold sometimes challenged his audience to find the limit of (sin tan x - tan sin x)/(asin atan x - atan asin x). This is hard to do using standard calculus techniques, but has a very simple geometric solution.

[@mathgeeks3598]

I use taylor expansion in mathematica and compute 7 times derivative for both numberator and denominator and verify the answer is one. Really impressive question

[@NovaWarrior77]

impressive answer

[@thomasblok2120]

I had a go using Taylor Series of both the numerator and the denominator, and the leading term in both is (-1/30)*x^7, so the limit is in fact 1. The working is a pain though, and I had to keep checking on wolfram alpha to correct my multiple errors.

[@viniciusmachadovogt]

Great explanation! Thank you!

[@stephenruby141]

Such a satisfying solution.

[@topdog5252]

What a funny story. Arnold said he went through 50 pages a day, in the video with him on the Shaw Prize. I wonder how many pages some people would waste on this question.

[@AlessioMortillaro]

What was the matematician's name cause I don't catch it?

[@user-ds4lg1zd5z]

Faltings, I think

[@accideux5182]

Gerd Faltings, Fields Medalist.

[@Nazliorcam]

❤️

[@stenzenneznets]

Little-o notation and Taylor series stopped at the first degree can be use to evaluate this limit.

[@alexanderchekmenev1217]

No, the first degree is not enough, because the first Taylor series terms cancel out each over and the first nonzero term is of degree 7: sin tan x - tan sin x = -1/30 x^7 + o(x^8)

[@thewatchman_returns]

Outstanding

[@user-yo7fi4uu2w]

Unfortunately the proof is wrong. The two functions lie on the same side with repect to y=x. You do not have the cross.

[@Nazliorcam]

❤️