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Limits Approaching Infinity

  1. Jun 26, 2011 #1
    1. The problem statement, all variables and given/known data
    How would you take this limit? The function is:
    [tex]lim_{x\rightarrow -∞}\frac{\sqrt{9x^{6}-x}}{x^{3}+1}[/tex]


    2. Relevant equations

    3. The attempt at a solution
    34oy2d2.jpg
    Uh, square root of negative infinity?

    Graphing tells me that this should go to -3.

    I just recalled that I can pull the 9 out to get 3root(-∞) ?
     
  2. jcsd
  3. Jun 26, 2011 #2

    eumyang

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    Nooooooo... you can't just divide by x3 when you have a square root in the numerator!
     
  4. Jun 26, 2011 #3

    gb7nash

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    You made an error with the squareroot. On the numerator, you're multiplying by 1/x^3 (which is fine). How do you proceed to move this into the squareroot?
     
  5. Jun 26, 2011 #4
    We briefly touched on this idea in class Friday, and only talked about polynomials, and the proof of the way we determine horizontal asymptotes without limits.

    [tex]lim_{x\rightarrow -∞}\frac{\frac{\sqrt{9x^{6}-x}}{x^3}}{1+\frac{1}{x^3}}[/tex]

    I'm really not sure where to go from there?
     
  6. Jun 26, 2011 #5

    eumyang

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    Note that
    [tex]x^3 = -\sqrt{x^6}[/tex]
    (since x is approaching negative infinity)

    So
    [tex]lim_{x\rightarrow -∞}\frac{\frac{\sqrt{9x^{6}-x}}{x^3}}{1+\frac{1}{x^3}} = lim_{x\rightarrow -∞}\frac{\frac{\sqrt{9x^{6}-x}}{-\sqrt{x^6}}}{1+\frac{1}{x^3}} = ...[/tex]
     
  7. Jun 26, 2011 #6
    20qark3.jpg

    I get what to do now, for the most part. Now I assume I must have some sign error. As noted above
    [tex]x^3=\sqrt{x^6}[/tex]
    I don't understand how you got the negative there. The exponent is odd. Is it because the limit to -inf assumes x is negative?
     
  8. Jun 26, 2011 #7

    Char. Limit

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    Yes, that's right. Since x is negative, x^3 is the negative square root of x^6.
     
  9. Jun 26, 2011 #8
    So, should I be dividing everything by the negative highest degree in the cases when x tends to negative infinity?

    If that is the case, then the top would evaluate to -3 sure, but the denominator would be -1 after the limit, making the solution positive 3?
     
  10. Jun 26, 2011 #9

    Char. Limit

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    No, no. You're still dividing top and bottom by x^3, not -x^3. However, in the "denominator of the numerator", you're going to change x^3 to -sqrt(x6). Now, why -sqrt(x6) and not +sqrt(x6)? The reason is because x^3 is only equal to sqrt(x6) for positive x. For negative x, x^3 is equal to -sqrt(x6).

    Hopefully I helped out a bit here.
     
  11. Jun 26, 2011 #10
    I'm not sure I understand why only the numerators denominator is getting the negative?
     
  12. Jun 26, 2011 #11

    SammyS

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    Because you leave the denominator's denominator as x3.

    In the numerator's denominator you change x3 to [itex]-\sqrt{x^6}\,.[/itex]
     
  13. Jun 26, 2011 #12
    So essentially, the whole expression is getting it's sign changed then?
     
  14. Jun 26, 2011 #13

    SammyS

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    Yes ... At least that's the way in appears.

    Actually, the expression is less than zero (i.e., negative) for x < -1 . It's just that you need to explicitly provide a negative sign when you change x3 to [itex]-\sqrt{x^6}[/itex] if x < 0 .
     
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