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Proof that sin(ax^2) ~ ax^2 as x->0 ?

Mathematics
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To make it look nicer, a proof of: \[\sin (ax^2)\sim ax^2, x \to 0\]
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
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, semi-rigorous algebraic proof of this... because I have no clue, how I would derive the former from the latter case.

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Other answers:

algebra will not do it because you have a trig function
Sorry, let me say it more precisely, a proof that does not require analysis, differentiation, or use of other, largely non-trivial methods, other than a) Geometrical arguments, b) Manipulation of the latter case, stated three posts ago. I hope I was more precise, now.
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sinx*cosx/2
areas of triangles
Yes, but what about sin(ax^2)?

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