anonymous
  • anonymous
Proof that sin(ax^2) ~ ax^2 as x->0 ?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
To make it look nicer, a proof of: \[\sin (ax^2)\sim ax^2, x \to 0\]
anonymous
  • 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
anonymous
  • 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.

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anonymous
  • anonymous
algebra will not do it because you have a trig function
anonymous
  • 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.
anonymous
  • anonymous
|dw:1346562207346:dw|
anonymous
  • anonymous
sinx*cosx/2
anonymous
  • anonymous
areas of triangles
anonymous
  • anonymous
Yes, but what about sin(ax^2)?

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