Question

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

Answer 1

To make it look nicer, a proof of:
\[\sin (ax^2)\sim ax^2, x \to 0\]

Answer 2

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

Answer 3

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.

Answer 4

algebra will not do it because you have a trig function

Answer 5

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.