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nurnabilashafiqah Group Title

lim sin^2 x/ 3x^2 x>0 using squeezing theorem .

  • one year ago
  • one year ago

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  1. Goten77 Group Title
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    |dw:1356618816515:dw|

    • one year ago
  2. akim0001 Group Title
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    1/3 ?

    • one year ago
  3. nurnabilashafiqah Group Title
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    yes ? help me to solve this question ..

    • one year ago
  4. Goten77 Group Title
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    well u can use locotoles rule |dw:1356631309095:dw|

    • one year ago
  5. Goten77 Group Title
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    |dw:1356631406328:dw|

    • one year ago
  6. Goten77 Group Title
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    |dw:1356631478538:dw|

    • one year ago
  7. Goten77 Group Title
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    |dw:1356631817966:dw||dw:1356631818210:dw|

    • one year ago
  8. nurnabilashafiqah Group Title
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    the 2nd last solution .. it's blank ?? the last equation ..how it can that ?

    • one year ago
  9. wio Group Title
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    @Goten77 "using squeezing theorem"

    • one year ago
  10. wio Group Title
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    How about?\[ \large \frac{-1}{ 3x^2} \leq \frac{\sin^2 (x)}{ 3x^2} \leq \frac{1}{ 3x^2} \]

    • one year ago
  11. wio Group Title
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    Hmm or maybe even \[ \large 0 \leq \frac{\sin^2 (x)}{ 3x^2} \leq \frac{1}{ 3x^2} \]

    • one year ago
  12. wio Group Title
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    Hmm, no, these won't work because they're not defined at 0...

    • one year ago
  13. nurnabilashafiqah Group Title
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    sin x . cos x become -sin^2 . cos^2 ??

    • one year ago
  14. nurnabilashafiqah Group Title
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    wio ?? can you help me .

    • one year ago
  15. wio Group Title
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    yes I am thinking about it.

    • one year ago
  16. wio Group Title
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    You need to find two functions... \(g(x)\) and \(f(x)\). The following must be true: \[ \lim_{x \to 0}g(x) = L \\ \lim_{x \to 0}f(x) = L \\ g(x) \leq \frac{\sin^2(x)}{3x^2} \leq f(x) \]

    • one year ago
  17. tkhunny Group Title
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    Have you considered \(\lim\limits_{x\rightarrow 0}\dfrac{sin(x)}{x} = 1\)?

    • one year ago
  18. nurnabilashafiqah Group Title
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    tkhunny : yes . I know about it . first rule in squeezing theorem, right ?

    • one year ago
  19. wio Group Title
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    Oh, I got an idea now...

    • one year ago
  20. tkhunny Group Title
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    Using that, we can find the limit directly. How do you suppose we inflict the Squeeze Theorem upon it?

    • one year ago
  21. nurnabilashafiqah Group Title
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    I don't got it .

    • one year ago
  22. RaphaelFilgueiras Group Title
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    |dw:1356929342251:dw|

    • one year ago
  23. nurnabilashafiqah Group Title
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    ouh . okay . now I get it ! thank you <3

    • one year ago
  24. tkhunny Group Title
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    How does \(\left(\dfrac{\sin(x)}{x}\right)^{2}\) compare to \(\dfrac{sin(x)}{x}\) for \(x \in (0,1)\)

    • one year ago
  25. nurnabilashafiqah Group Title
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    just separate it, right ? @RaphaelFilgueiras, @wio, @tkhunny

    • one year ago
  26. wio Group Title
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    \(x\in [-1, 1] \)

    • one year ago
  27. wio Group Title
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    \(\epsilon > 0,\quad x\in [0-\epsilon, 0 + \epsilon] \)

    • one year ago
  28. RaphaelFilgueiras Group Title
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    x must be in radians,this is the only restriction to aply the fundamental trygonometric limit

    • one year ago
  29. nurnabilashafiqah Group Title
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    thanks !

    • one year ago
  30. wio Group Title
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    @nurnabilashafiqah That's not using the squeeze theorem, so it's not what the question is asking of you.

    • one year ago
  31. wio Group Title
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    You need to find two functions that are equal to 1/3 at 0, one function must be greater than your function, the other must be less than your function

    • one year ago
  32. wio Group Title
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    Use something like the fact that \[ \sin^2(x) \leq x^2 \]

    • one year ago
  33. wio Group Title
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    That can get you an upperbound. Then you need a lower bound.

    • one year ago
  34. RaphaelFilgueiras Group Title
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    @wio is rigth i did not see the part of squeezing

    • one year ago
  35. sirm3d Group Title
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    \[\large \cos x < \frac{\sin x}{x} < \frac{1}{\cos x}\]

    • one year ago
  36. RaphaelFilgueiras Group Title
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    but if you are in radian you will have 0 < sinx²<x²

    • one year ago
  37. nurnabilashafiqah Group Title
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    squeezing theorem have both less than and greater than ? sorry, I'm new about this topic . please explain to me briefly .

    • one year ago
  38. RaphaelFilgueiras Group Title
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    |dw:1356930269595:dw|

    • one year ago
  39. RaphaelFilgueiras Group Title
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    that arc is x because in radian system we use 1 as radius

    • one year ago
  40. wio Group Title
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    Okay so I'm thinking the best to use is: \[ \frac{\cos^2(x) }{3} \leq \frac{\sin^2(x)}{3x^2} \leq \frac{x^2}{3x^2} \]

    • one year ago
  41. RaphaelFilgueiras Group Title
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    so we have sinx<x<tgx

    • one year ago
  42. nurnabilashafiqah Group Title
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    huh ?

    • one year ago
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