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LolWolfBest ResponseYou've already chosen the best response.0
To make it look nicer, a proof of: \[\sin (ax^2)\sim ax^2, x \to 0\]
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
same as \[\lim_{x\to 0}\frac{\sin(ax^2)}{ax^2}=1\] many methods but if you know that \[\lim_{x\to 0}\frac{\sin(x)}{x}=1\] then this is identical
 one year ago

LolWolfBest ResponseYou've already chosen the best response.0
Well, a proof without l'Hospitals rule, I mean, I can prove the rule and I can prove the latter case trigonometrically and using IVT... but I want a nice, semirigorous algebraic proof of this... because I have no clue, how I would derive the former from the latter case.
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
algebra will not do it because you have a trig function
 one year ago

LolWolfBest ResponseYou've already chosen the best response.0
Sorry, let me say it more precisely, a proof that does not require analysis, differentiation, or use of other, largely nontrivial methods, other than a) Geometrical arguments, b) Manipulation of the latter case, stated three posts ago. I hope I was more precise, now.
 one year ago

cinarBest ResponseYou've already chosen the best response.0
sinx*cosx/2<sinx/2<tanx/2
 one year ago

LolWolfBest ResponseYou've already chosen the best response.0
Yes, but what about sin(ax^2)?
 one year ago
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