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LolWolf Group Title

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

  • 2 years ago
  • 2 years ago

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

    • 2 years ago
  2. satellite73 Group Title
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    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

    • 2 years ago
  3. LolWolf Group Title
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    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.

    • 2 years ago
  4. satellite73 Group Title
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    algebra will not do it because you have a trig function

    • 2 years ago
  5. LolWolf Group Title
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    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.

    • 2 years ago
  6. cinar Group Title
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    |dw:1346562207346:dw|

    • 2 years ago
  7. cinar Group Title
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    sinx*cosx/2<sinx/2<tanx/2

    • 2 years ago
  8. cinar Group Title
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    areas of triangles

    • 2 years ago
  9. LolWolf Group Title
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    Yes, but what about sin(ax^2)?

    • 2 years ago
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