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Littlebird

  • one year ago

lim x-> -infinity sqrt(4x^6-x)/(x^3+5)

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  1. Littlebird
    • one year ago
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    @DanJS

  2. Littlebird
    • one year ago
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    This square root is messing me up.

  3. DanJS
    • one year ago
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    square root of that ratio, or just root of numerator

  4. Littlebird
    • one year ago
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    of the numerator

  5. DanJS
    • one year ago
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    similar to tha tlast one, i think , ...

  6. DanJS
    • one year ago
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    take as much of the x terms out of the root as you can

  7. DanJS
    • one year ago
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    try and manipulate that root, so that an infinit x value is in the denominator of a term... so it goes away... maybe

  8. Littlebird
    • one year ago
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    |dw:1443228681001:dw|

  9. Littlebird
    • one year ago
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    that mess is basically one of my attempts, but it looks like I'm going to get a positive result. wolfram says it is -2, but it won't explain why

  10. Littlebird
    • one year ago
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    I also tried conjugating it, but that also looked insane

  11. DanJS
    • one year ago
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    you can't take the root of a quantity with addition like that

  12. DanJS
    • one year ago
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    root(x+a) is not same as root(x) + root (a)

  13. Littlebird
    • one year ago
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    I sort of knew that, but I was willing to try mathematical blasphemy to solve this

  14. DanJS
    • one year ago
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    maybe try dividing through by root(x^6)

  15. DanJS
    • one year ago
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    the numerator and denominator

  16. DanJS
    • one year ago
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    \[\Large \frac{ [\sqrt{4x^6-x}]*\frac{ 1 }{ \sqrt{x^6} } }{[ x^3 + 5]*\frac{ 1 }{ \sqrt{x^6} } }\]

  17. DanJS
    • one year ago
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    seems promising

  18. DanJS
    • one year ago
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    trying to get a constant - a fraction over x in the root

  19. DanJS
    • one year ago
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    and since it is a book prob, root(x^6) = x^3 yay

  20. DanJS
    • one year ago
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    k, so try reorganizing things

  21. DanJS
    • one year ago
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    distribute, combine roots, simplify

  22. DanJS
    • one year ago
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    you can't just divide the denominator by something, you have to do the same to everything else, the numerator

  23. DanJS
    • one year ago
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    4/6 not same as 4/(6/a)

  24. DanJS
    • one year ago
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    have to divide everything

  25. Littlebird
    • one year ago
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    do I conjugate?

  26. DanJS
    • one year ago
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    just reorganize to simplify, ill start typing the next piece, remember a root over a root is the same as the root of the fraction together

  27. DanJS
    • one year ago
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    \[\sqrt{\frac{ a }{ b }}= \frac{ \sqrt{a} }{ \sqrt{b} }\]

  28. DanJS
    • one year ago
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    For the top, and for the bottom just distribute and divide if you can

  29. Littlebird
    • one year ago
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    i now think I did the division wrong earlier

  30. DanJS
    • one year ago
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    \[\Large \frac{ [\sqrt{4x^6-x}]*\frac{ 1 }{ \sqrt{x^6} } }{[ x^3 + 5]*\frac{ 1 }{ \sqrt{x^6} } }~=~\frac{ \sqrt{\frac{ 4x^6-x }{ x^6 }} }{ \frac{ x^3 }{ \sqrt{x^6}}+\frac{ 5 }{ \sqrt{x^6} } }\]

  31. DanJS
    • one year ago
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    let me know if you dont see how everything worked

  32. DanJS
    • one year ago
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    \[\large = \frac{ \sqrt{4-\frac{ 1 }{ x^5 }} }{ 1+ \frac{ 5 }{ x^3 } }\]

  33. Littlebird
    • one year ago
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    I feel silly. I just realized now what you meant by dividing by sqrt(x^6). I was still trying to divide by x^3

  34. DanJS
    • one year ago
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    or x^3, they are the same

  35. DanJS
    • one year ago
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    root of x^6 is x^3

  36. DanJS
    • one year ago
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    nice book prob;lem.. hah

  37. DanJS
    • one year ago
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    \[\large \sqrt{x^6} = [x^6]^{1/2} = x^{6/2} = x^3\]

  38. DanJS
    • one year ago
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    yeah, was trying to find something to divide by that would leave a number + a fraction over x under the root

  39. DanJS
    • one year ago
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    after this is done, i think i remember an even better way when the limt is to infinity..

  40. Littlebird
    • one year ago
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    |dw:1443231058854:dw|

  41. Littlebird
    • one year ago
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    I know I'm missing something, since the answer is negative for some reason

  42. Littlebird
    • one year ago
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    Is it because when I divided by sqrt(x^6) I am technically supposed to be dividing by -x^3 since x->-infinity is negative?

  43. DanJS
    • one year ago
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    umm, sorry was afk

  44. DanJS
    • one year ago
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    bad though, forget that

  45. Littlebird
    • one year ago
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    ok

  46. DanJS
    • one year ago
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    property 6 on that pdf

  47. DanJS
    • one year ago
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    then property 2, it is good

  48. Littlebird
    • one year ago
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    where did the negative sign come from?

  49. DanJS
    • one year ago
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    oh i think i see how it will be negative

  50. DanJS
    • one year ago
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    i think it has to do with the square root property, a square root is always a positive value \[\huge \sqrt{[f(x)]^2} = \left| f(x) \right|\]

  51. DanJS
    • one year ago
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    something to do with an odd power and or negative infinity value, and that root property, lol i can't think

  52. Littlebird
    • one year ago
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    I will try to Google your words

  53. DanJS
    • one year ago
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    if you pull out a root x^2 = abs(x) from the initial numorater, that will change the sign i think, just different solving

  54. DanJS
    • one year ago
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    or probably something simple i just cant think of to fix this as is

  55. DanJS
    • one year ago
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    since the limit is to negative infinity, you can assume the x value is negative...

  56. DanJS
    • one year ago
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    look through example 3 on this page, it is similar and probably has the answer in it http://tutorial.math.lamar.edu/Classes/CalcI/LimitsAtInfinityI.aspx

  57. Littlebird
    • one year ago
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    I saw it. It looks like I have another obscure rule to remember

  58. DanJS
    • one year ago
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    if you have an abs val of x, and the limit is to infinity, then i assume you can call x negative, so the absolute value will be -x

  59. DanJS
    • one year ago
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    neg infinity limit

  60. DanJS
    • one year ago
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    \[\left| x \right| = x,~~if ~x \ge 0~~and~~~\left| x \right|=-x~~if~x<0\] that is probably where the change in sign comes from

  61. DanJS
    • one year ago
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    @jim_thompson5910

  62. DanJS
    • one year ago
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    oh, the other thing i think you can do is just forget about the x in the top and the 5 in the bottom , since they are magnitudes less effective for large x values to infinity

  63. DanJS
    • one year ago
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    than the other parts of the quantities

  64. Littlebird
    • one year ago
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    That does work, but I still have to remember the negative

  65. DanJS
    • one year ago
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    \[\large \lim_{x \rightarrow -\infty} [\frac{ \sqrt{4x^6} }{ x^3 }]\]

  66. DanJS
    • one year ago
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    i see, that works using the abs value function...

  67. DanJS
    • one year ago
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    \[\frac{ 2\sqrt{x^6} }{ x^3 }\]

  68. DanJS
    • one year ago
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    \[\large \frac{ 2\sqrt{[x^3]^2} }{ x^3 }\]

  69. Littlebird
    • one year ago
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    Thanks for the help! If I stop replying, it's because of a magic show in 10 min

  70. DanJS
    • one year ago
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    i think that works, use the absolute value function, the sign should change

  71. Littlebird
    • one year ago
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    I will remeber both ways to be safe

  72. DanJS
    • one year ago
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    lol , calc is 99.9% algebra with a small initial shortcut from calc

  73. DanJS
    • one year ago
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    \[\large \frac{ 2*\left| x \right|^3 }{ x^3 } \]

  74. DanJS
    • one year ago
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    since the limit is out to negative infinity, you can assume that the x value will be negative From, the absolute value function, \[\left| x \right| = -x~~~if~~~x<0\]

  75. DanJS
    • one year ago
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    \[\huge \frac{ 2*(-x)^3 }{ x^3 }\]

  76. DanJS
    • one year ago
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    \[\huge \frac{ 2*(-1)^3(x^3 )}{ x^3 } = -2\] yeah, sounds good to me, just wonder what needed to be done, to the first manipulation to put in that sign change...

  77. DanJS
    • one year ago
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    \[\Large \frac{ [\sqrt{4x^6-x}]*\frac{ 1 }{ \sqrt{x^6} } }{[ x^3 + 5]*\frac{ 1 }{ \sqrt{x^6} } } = \frac{ \sqrt{4-\frac{ 1 }{ x^5 }} }{ 1+ \frac{ 5 }{ x^3 } } \] i am done here for now

  78. DanJS
    • one year ago
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    figured what to do, same reason as the solution above, just use the absolute value function

  79. DanJS
    • one year ago
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    \[\Large \frac{ [\sqrt{4x^6-x}]*\frac{ 1 }{ \sqrt{x^6} } }{[ x^3 + 5]*\frac{ 1 }{ \sqrt{x^6} } }~=~\frac{ \sqrt{\frac{ 4x^6-x }{ x^6 }} }{ \frac{ x^3 }{ \sqrt{x^6}}+\frac{ 5 }{ \sqrt{x^6} } }\]

  80. DanJS
    • one year ago
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    \[\large =\frac{ \sqrt{4-1/x^5} }{ \frac{ x^3 }{ \left| x^3 \right| } +\frac{ 5 }{ \left| x^3 \right| }}\]

  81. DanJS
    • one year ago
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    same logic as the solution above, limit to negative infinity means a negative x value, so the absolute value (x) is -x, and it stays negative for an odd power of 3, there is the change in friggin sign,

  82. DanJS
    • one year ago
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    limit value at -infinity = 2 / -1 = - 2

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