## Littlebird one year ago lim x-> -infinity sqrt(4x^6-x)/(x^3+5)

1. Littlebird

@DanJS

2. Littlebird

This square root is messing me up.

3. DanJS

square root of that ratio, or just root of numerator

4. Littlebird

of the numerator

5. DanJS

similar to tha tlast one, i think , ...

6. DanJS

take as much of the x terms out of the root as you can

7. DanJS

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

|dw:1443228681001:dw|

9. Littlebird

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

I also tried conjugating it, but that also looked insane

11. DanJS

you can't take the root of a quantity with addition like that

12. DanJS

root(x+a) is not same as root(x) + root (a)

13. Littlebird

I sort of knew that, but I was willing to try mathematical blasphemy to solve this

14. DanJS

maybe try dividing through by root(x^6)

15. DanJS

the numerator and denominator

16. DanJS

$\Large \frac{ [\sqrt{4x^6-x}]*\frac{ 1 }{ \sqrt{x^6} } }{[ x^3 + 5]*\frac{ 1 }{ \sqrt{x^6} } }$

17. DanJS

seems promising

18. DanJS

trying to get a constant - a fraction over x in the root

19. DanJS

and since it is a book prob, root(x^6) = x^3 yay

20. DanJS

k, so try reorganizing things

21. DanJS

distribute, combine roots, simplify

22. DanJS

you can't just divide the denominator by something, you have to do the same to everything else, the numerator

23. DanJS

4/6 not same as 4/(6/a)

24. DanJS

have to divide everything

25. Littlebird

do I conjugate?

26. DanJS

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

$\sqrt{\frac{ a }{ b }}= \frac{ \sqrt{a} }{ \sqrt{b} }$

28. DanJS

For the top, and for the bottom just distribute and divide if you can

29. Littlebird

i now think I did the division wrong earlier

30. DanJS

$\Large \frac{ [\sqrt{4x^6-x}]*\frac{ 1 }{ \sqrt{x^6} } }{[ x^3 + 5]*\frac{ 1 }{ \sqrt{x^6} } }~=~\frac{ \sqrt{\frac{ 4x^6-x }{ x^6 }} }{ \frac{ x^3 }{ \sqrt{x^6}}+\frac{ 5 }{ \sqrt{x^6} } }$

31. DanJS

let me know if you dont see how everything worked

32. DanJS

$\large = \frac{ \sqrt{4-\frac{ 1 }{ x^5 }} }{ 1+ \frac{ 5 }{ x^3 } }$

33. Littlebird

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

or x^3, they are the same

35. DanJS

root of x^6 is x^3

36. DanJS

nice book prob;lem.. hah

37. DanJS

$\large \sqrt{x^6} = [x^6]^{1/2} = x^{6/2} = x^3$

38. DanJS

yeah, was trying to find something to divide by that would leave a number + a fraction over x under the root

39. DanJS

after this is done, i think i remember an even better way when the limt is to infinity..

40. Littlebird

|dw:1443231058854:dw|

41. Littlebird

I know I'm missing something, since the answer is negative for some reason

42. Littlebird

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

umm, sorry was afk

44. DanJS

45. Littlebird

ok

46. DanJS

property 6 on that pdf

47. DanJS

then property 2, it is good

48. Littlebird

where did the negative sign come from?

49. DanJS

oh i think i see how it will be negative

50. DanJS

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

something to do with an odd power and or negative infinity value, and that root property, lol i can't think

52. Littlebird

53. DanJS

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

or probably something simple i just cant think of to fix this as is

55. DanJS

since the limit is to negative infinity, you can assume the x value is negative...

56. DanJS

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

I saw it. It looks like I have another obscure rule to remember

58. DanJS

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

neg infinity limit

60. DanJS

$\left| x \right| = x,~~if ~x \ge 0~~and~~~\left| x \right|=-x~~if~x<0$ that is probably where the change in sign comes from

61. DanJS

@jim_thompson5910

62. DanJS

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

than the other parts of the quantities

64. Littlebird

That does work, but I still have to remember the negative

65. DanJS

$\large \lim_{x \rightarrow -\infty} [\frac{ \sqrt{4x^6} }{ x^3 }]$

66. DanJS

i see, that works using the abs value function...

67. DanJS

$\frac{ 2\sqrt{x^6} }{ x^3 }$

68. DanJS

$\large \frac{ 2\sqrt{[x^3]^2} }{ x^3 }$

69. Littlebird

Thanks for the help! If I stop replying, it's because of a magic show in 10 min

70. DanJS

i think that works, use the absolute value function, the sign should change

71. Littlebird

I will remeber both ways to be safe

72. DanJS

lol , calc is 99.9% algebra with a small initial shortcut from calc

73. DanJS

$\large \frac{ 2*\left| x \right|^3 }{ x^3 }$

74. DanJS

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

$\huge \frac{ 2*(-x)^3 }{ x^3 }$

76. DanJS

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

77. DanJS

$\Large \frac{ [\sqrt{4x^6-x}]*\frac{ 1 }{ \sqrt{x^6} } }{[ x^3 + 5]*\frac{ 1 }{ \sqrt{x^6} } } = \frac{ \sqrt{4-\frac{ 1 }{ x^5 }} }{ 1+ \frac{ 5 }{ x^3 } }$ i am done here for now

78. DanJS

figured what to do, same reason as the solution above, just use the absolute value function

79. DanJS

$\Large \frac{ [\sqrt{4x^6-x}]*\frac{ 1 }{ \sqrt{x^6} } }{[ x^3 + 5]*\frac{ 1 }{ \sqrt{x^6} } }~=~\frac{ \sqrt{\frac{ 4x^6-x }{ x^6 }} }{ \frac{ x^3 }{ \sqrt{x^6}}+\frac{ 5 }{ \sqrt{x^6} } }$

80. DanJS

$\large =\frac{ \sqrt{4-1/x^5} }{ \frac{ x^3 }{ \left| x^3 \right| } +\frac{ 5 }{ \left| x^3 \right| }}$

81. DanJS

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

limit value at -infinity = 2 / -1 = - 2