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anonymous
 4 years ago
Proof that sin(ax^2) ~ ax^2 as x>0 ?
anonymous
 4 years ago
Proof that sin(ax^2) ~ ax^2 as x>0 ?

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0same 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Well, 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0algebra will not do it because you have a trig function

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Sorry, 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1346562207346:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0sinx*cosx/2<sinx/2<tanx/2

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yes, but what about sin(ax^2)?
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