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roselin

  • 2 years ago

whats the lim x goes to infinity of 2/square root of x

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  1. JoãoVitorMC
    • 2 years ago
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    0

  2. roselin
    • 2 years ago
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    yes

  3. zepdrix
    • 2 years ago
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    Rose if you're not convinced, just plug a large number into your calculator to check! :) if you plug one bajillion in for x, what value do you get for the limit? \[\frac{ 2 }{ \sqrt{bajillion} }\approx \frac{ 2 }{ 999999999 } \approx 0\]

  4. JoãoVitorMC
    • 2 years ago
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    yeah, but do what zepdrix said only for check, because if you do this in a test for example, it will be wrong because infinity is NOT a number...

  5. roselin
    • 2 years ago
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    okay

  6. roselin
    • 2 years ago
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    it is always 0 ?

  7. JoãoVitorMC
    • 2 years ago
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    no! it will depends on what you are taking the limit, always try to simplify the limit, try to do not have a denominator, in this case we can not simplify so we do this

  8. roselin
    • 2 years ago
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    i am trying to solve this problem where x goes to infinity 2+square root of x/ 2- square root of x

  9. roselin
    • 2 years ago
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    i am getting the answer as 1, is it right?

  10. JoãoVitorMC
    • 2 years ago
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    \[\lim_{x \rightarrow \infty}\frac{2+\sqrt{x}}{2-\sqrt{x}}\]

  11. JoãoVitorMC
    • 2 years ago
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    this one?

  12. roselin
    • 2 years ago
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    yes,

  13. JoãoVitorMC
    • 2 years ago
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    the limit is equal to -1

  14. JoãoVitorMC
    • 2 years ago
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    have you already study L'hospital rule ?

  15. roselin
    • 2 years ago
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    can you please explain how you got -1?

  16. roselin
    • 2 years ago
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    No,

  17. JoãoVitorMC
    • 2 years ago
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    ok let me explain

  18. JoãoVitorMC
    • 2 years ago
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    |dw:1350703643878:dw|

  19. JoãoVitorMC
    • 2 years ago
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    got it?

  20. roselin
    • 2 years ago
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    what did you do with the whole numerator thing? u ended up with X

  21. JoãoVitorMC
    • 2 years ago
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    when x goes to infinity, the other terms with lower power we take them off, and stay only with the higher terms them simply if possible and aply the limit

  22. JoãoVitorMC
    • 2 years ago
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    i will post how i got the numerator

  23. roselin
    • 2 years ago
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    ohh , u mean the limit of 4 and 4 sqrt x = 0?

  24. JoãoVitorMC
    • 2 years ago
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    |dw:1350704041192:dw|

  25. JoãoVitorMC
    • 2 years ago
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    no this i simply take them out because comparing with infinity they make no difference

  26. roselin
    • 2 years ago
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    oh okay,

  27. JoãoVitorMC
    • 2 years ago
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    when you deal with limits on the infinity you can make this assumptions

  28. JoãoVitorMC
    • 2 years ago
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    but take care...

  29. JoãoVitorMC
    • 2 years ago
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    when you see L'hostpital's Rule this limits will became very easy to solve.... you will see

  30. roselin
    • 2 years ago
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    we have not gotten to that yet.

  31. JoãoVitorMC
    • 2 years ago
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    yeah

  32. roselin
    • 2 years ago
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    if you do not mind then, I have one more problem. would you explain that one?

  33. JoãoVitorMC
    • 2 years ago
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    sure

  34. roselin
    • 2 years ago
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    thank you

  35. roselin
    • 2 years ago
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    lim sec x x goes to( -pi/2)^+

  36. JoãoVitorMC
    • 2 years ago
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    let me solve here one moment

  37. roselin
    • 2 years ago
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    okay

  38. JoãoVitorMC
    • 2 years ago
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    it is equal to infinity, i will draw here

  39. roselin
    • 2 years ago
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    okay

  40. JoãoVitorMC
    • 2 years ago
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    |dw:1350704691784:dw|

  41. JoãoVitorMC
    • 2 years ago
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    is it clear?

  42. roselin
    • 2 years ago
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    yes,viewing it.

  43. JoãoVitorMC
    • 2 years ago
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    see, when we have something divide by a number really really really small them it becomes a really really really large number so the limit tends to the infinity

  44. roselin
    • 2 years ago
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    okay, but how did you get that 10?

  45. JoãoVitorMC
    • 2 years ago
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    it is just a notation for this: 0.00000000000000000000000...00000000000001, then we represent this really small or large number with 10 power

  46. roselin
    • 2 years ago
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    oh ,okay. got it.

  47. roselin
    • 2 years ago
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    wait,wait

  48. roselin
    • 2 years ago
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    that (-pi/2)^+ is the same as (-pi/2)?

  49. JoãoVitorMC
    • 2 years ago
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    ALMOST the same, when we have this + is a little more than pi/2

  50. JoãoVitorMC
    • 2 years ago
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    -pi/2, sorry

  51. roselin
    • 2 years ago
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    so, the asnwer will be the same one right?

  52. roselin
    • 2 years ago
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    *answer

  53. JoãoVitorMC
    • 2 years ago
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    no, because if we take 1/0 it is not allowed. But 1/0.00000000001 is allowed

  54. JoãoVitorMC
    • 2 years ago
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    we can not divide by 0, NEVER

  55. roselin
    • 2 years ago
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    yes,that is true.

  56. JoãoVitorMC
    • 2 years ago
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    so you got it?

  57. roselin
    • 2 years ago
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    yeah.

  58. roselin
    • 2 years ago
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    Thank you soooo much.

  59. JoãoVitorMC
    • 2 years ago
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    yw :)

  60. roselin
    • 2 years ago
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    good night, byee

  61. JoãoVitorMC
    • 2 years ago
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    good, bye

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