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nurnabilashafiqah

  • 3 years ago

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

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  1. Goten77
    • 3 years ago
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    |dw:1356618816515:dw|

  2. akim0001
    • 3 years ago
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    1/3 ?

  3. nurnabilashafiqah
    • 3 years ago
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    yes ? help me to solve this question ..

  4. Goten77
    • 3 years ago
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    well u can use locotoles rule |dw:1356631309095:dw|

  5. Goten77
    • 3 years ago
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    |dw:1356631406328:dw|

  6. Goten77
    • 3 years ago
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    |dw:1356631478538:dw|

  7. Goten77
    • 3 years ago
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    |dw:1356631817966:dw||dw:1356631818210:dw|

  8. nurnabilashafiqah
    • 3 years ago
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    the 2nd last solution .. it's blank ?? the last equation ..how it can that ?

  9. wio
    • 3 years ago
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    @Goten77 "using squeezing theorem"

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

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

  12. wio
    • 3 years ago
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    Hmm, no, these won't work because they're not defined at 0...

  13. nurnabilashafiqah
    • 3 years ago
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    sin x . cos x become -sin^2 . cos^2 ??

  14. nurnabilashafiqah
    • 3 years ago
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    wio ?? can you help me .

  15. wio
    • 3 years ago
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    yes I am thinking about it.

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

  17. tkhunny
    • 3 years ago
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    Have you considered \(\lim\limits_{x\rightarrow 0}\dfrac{sin(x)}{x} = 1\)?

  18. nurnabilashafiqah
    • 3 years ago
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    tkhunny : yes . I know about it . first rule in squeezing theorem, right ?

  19. wio
    • 3 years ago
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    Oh, I got an idea now...

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

  21. nurnabilashafiqah
    • 3 years ago
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    I don't got it .

  22. RaphaelFilgueiras
    • 3 years ago
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    |dw:1356929342251:dw|

  23. nurnabilashafiqah
    • 3 years ago
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    ouh . okay . now I get it ! thank you <3

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

  25. nurnabilashafiqah
    • 3 years ago
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    just separate it, right ? @RaphaelFilgueiras, @wio, @tkhunny

  26. wio
    • 3 years ago
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    \(x\in [-1, 1] \)

  27. wio
    • 3 years ago
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    \(\epsilon > 0,\quad x\in [0-\epsilon, 0 + \epsilon] \)

  28. RaphaelFilgueiras
    • 3 years ago
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    x must be in radians,this is the only restriction to aply the fundamental trygonometric limit

  29. nurnabilashafiqah
    • 3 years ago
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    thanks !

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

  31. wio
    • 3 years ago
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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

  32. wio
    • 3 years ago
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    Use something like the fact that \[ \sin^2(x) \leq x^2 \]

  33. wio
    • 3 years ago
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    That can get you an upperbound. Then you need a lower bound.

  34. RaphaelFilgueiras
    • 3 years ago
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    @wio is rigth i did not see the part of squeezing

  35. sirm3d
    • 3 years ago
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    \[\large \cos x < \frac{\sin x}{x} < \frac{1}{\cos x}\]

  36. RaphaelFilgueiras
    • 3 years ago
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    but if you are in radian you will have 0 < sinx²<x²

  37. nurnabilashafiqah
    • 3 years ago
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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 .

  38. RaphaelFilgueiras
    • 3 years ago
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    |dw:1356930269595:dw|

  39. RaphaelFilgueiras
    • 3 years ago
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    that arc is x because in radian system we use 1 as radius

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

  41. RaphaelFilgueiras
    • 3 years ago
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    so we have sinx<x<tgx

  42. nurnabilashafiqah
    • 3 years ago
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    huh ?

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