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

1. anonymous

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

2. anonymous

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

3. anonymous

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.

4. anonymous

algebra will not do it because you have a trig function

5. anonymous

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.

6. anonymous

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7. anonymous

sinx*cosx/2<sinx/2<tanx/2

8. anonymous

areas of triangles

9. anonymous

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