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Christos

  • one year ago

Can you help me find this limit: http://screencast.com/t/6CD9ErAG3N9m All I need is the first step, I am gonna climb from there easily. Sorry if it's a little unclear picture, it's from an old book

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  1. Christos
    • one year ago
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    @ganeshie8 @Mertsj

  2. robz8
    • one year ago
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    for reference, the answer is \[-\sqrt3/2\] i do not remember the steps though, sorry

  3. Christos
    • one year ago
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    I know the answer too my friend, but I dont know how to find it, thanks though

  4. robz8
    • one year ago
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    sorry i could not help :/

  5. Stewie_Griffin_
    • one year ago
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    CANT EVEN SEE THE QUESTION

  6. Christos
    • one year ago
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    How Come?? The guy above you saw it , I posted a picture

  7. Jhannybean
    • one year ago
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    I can't see it either. what is the limit going to? Can't make it out after the arrow

  8. Christos
    • one year ago
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    infinity+ ....

  9. Christos
    • one year ago
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    + infinity ***** I am sorry

  10. robz8
    • one year ago
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    lim of x going to positive infinity of sin((pi*x)/(2-3x))

  11. Jhannybean
    • one year ago
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    \[\lim_{x \rightarrow +\infty}\sin (\frac{ \pi x }{ 2-3x })\]

  12. Stewie_Griffin_
    • one year ago
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    can you actually take the pic clearly?

  13. rajee_sam
    • one year ago
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    \[\lim_{x \rightarrow +\infty} Sin (\frac{ \pi x }{ 2-3x })\]

  14. Christos
    • one year ago
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    Yes

  15. Jhannybean
    • one year ago
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    Now Stewie can stop trolling, lol. xD

  16. Stewie_Griffin_
    • one year ago
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    VICTORY?

  17. rajee_sam
    • one year ago
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    I want to learn this too

  18. Jhannybean
    • one year ago
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    Same here.

  19. Stewie_Griffin_
    • one year ago
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    yes or no?

  20. rajee_sam
    • one year ago
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    who is teaching??

  21. Jhannybean
    • one year ago
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    You can divide by the highest power in the denominator which is x

  22. Christos
    • one year ago
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    But how does that lead you to the solution which is -sqrt(3)/2

  23. Jhannybean
    • one year ago
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    Can't you? \[\large \lim_{x \rightarrow +\infty}\sin(\frac{ \pi x/x }{ \frac{ 2 }{ x }-\frac{ 3x }{ x } })\]

  24. Christos
    • one year ago
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    yes and then?

  25. Jhannybean
    • one year ago
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    So what is \[\sin(-\frac{ \pi }{ 3 })\]? Whcih quadrant is sine negative ?

  26. rajee_sam
    • one year ago
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    \[\lim_{x \rightarrow +\infty} \sin \frac{ x }{ x }(\frac{ \pi }{ \frac{ 2 }{ x } - 3 })\]

  27. Christos
    • one year ago
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    @myko @timo86m @jim_thompson5910 @jhonyy9 @Hero @Euler271

  28. Christos
    • one year ago
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    The result is most likely engative yea

  29. rajee_sam
    • one year ago
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    Sin (- pi / 3)

  30. Christos
    • one year ago
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    negative*

  31. Christos
    • one year ago
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    yes but HOW @rajee_sam

  32. Jhannybean
    • one year ago
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    Sine is negative in quadrant 3 and 4, so if \[\sin \frac{ \pi }{ 3 } = \frac{ \sqrt{3} }{ 2 }\] then \[\sin -\frac{ \pi }{ 3 }= -\frac{ \sqrt{3} }{ 2 }\]

  33. Christos
    • one year ago
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    Am I eligible to differantiate this thing ? because if we differantiate we get sin(pi/-3)

  34. Christos
    • one year ago
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    instantly

  35. rajee_sam
    • one year ago
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    |dw:1369516434913:dw|

  36. Jhannybean
    • one year ago
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    You just need to find the positive radian measure, which is \[\frac{ \pi }{ }\] and then translate it accordingly to find which negative value is correlated with the positive radian measure

  37. rajee_sam
    • one year ago
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    \[\sin ( -\theta ) = - \sin (\theta)\]

  38. Jhannybean
    • one year ago
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    sorry,i meant pi/3 *

  39. Christos
    • one year ago
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    how did you get the x out @rajee_sam

  40. rajee_sam
    • one year ago
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    I am taking a GCF for both numerator and denominator so that I can cancel them out and do not have any x multiplied with anything

  41. Christos
    • one year ago
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    GCF ?

  42. rajee_sam
    • one year ago
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    Common Factor

  43. phi
    • one year ago
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    \[\lim_{x \rightarrow +\infty}\sin \left(\frac{ \pi x }{ 2-3x }\right) = \sin\left( \lim_{x \rightarrow +\infty} \frac{ \pi x }{ 2-3x }\right)\] as noted above you can rewrite \[ \frac{ \pi x }{ 2-3x } = \frac{ \pi }{\frac{2}{x} -3} \]

  44. Christos
    • one year ago
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    Alright! Alright! I solved it this way! By during the time I was trying to solve this I figured out a second way! Tell me if this is correct: Just taking the derivative of both denominator and numerator and then we are done. Can I apply this to ANY limit? Because here it works!

  45. Jhannybean
    • one year ago
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    that's what i did pi.... :(

  46. Christos
    • one year ago
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    @phi

  47. Jhannybean
    • one year ago
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    if you're taking the derivative you'll have to use chain rule on the whole thing, sin (whatever's inside) and then * whatever is inside.

  48. phi
    • one year ago
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    yes, you can use L'Hopital's rule http://en.wikipedia.org/wiki/L'Hôpital's_rule if you get 0/0 or inf/inf

  49. Christos
    • one year ago
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    and no chain rule??

  50. Christos
    • one year ago
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    I am talking about this specific problem

  51. phi
    • one year ago
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    no, use chain rule. use \[ \lim_{x \rightarrow +\infty}\sin \left(\frac{ \pi x }{ 2-3x }\right) = \sin\left( \lim_{x \rightarrow +\infty} \frac{ \pi x }{ 2-3x }\right) \]

  52. phi
    • one year ago
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    *do not use chain rule

  53. Christos
    • one year ago
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    but even if I dont use the chain rule we end up to the same result

  54. Christos
    • one year ago
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    Ohh so this formula you showed to me just now is only when we have 0/0 or infinite/infinite ?

  55. Jhannybean
    • one year ago
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    Oh I see, and you'd use LH rule to simplify it further and end up with the same result.

  56. phi
    • one year ago
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    using L'Hopital's rule needs 0/0 or inf/inf use it on \[ \lim_{x \rightarrow +\infty} \frac{ \pi x }{ 2-3x } \]

  57. Christos
    • one year ago
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    ok and I get this http://screencast.com/t/jPlMQFrZ What now?

  58. phi
    • one year ago
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    you can use L'Hopital d top/ d bottom = pi/-3 or you can use algebra

  59. Christos
    • one year ago
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    and the lim(x-->+infinity) goes away?

  60. Christos
    • one year ago
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    oh I get you now

  61. Christos
    • one year ago
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    but still I am stuck on the lim thingy

  62. phi
    • one year ago
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    then sin (-pi/3) = -sqrt{3}/2 *** and the lim(x-->+infinity) goes away? if you use L'Hopital \[ \lim_{x \rightarrow +\infty} \frac{ \pi x }{ 2-3x } = \lim_{x \rightarrow +\infty} \frac{\pi}{-3}= -\frac{\pi}{3}\]

  63. Christos
    • one year ago
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    I see, thank you all

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