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Limits to Infinity

  1. Jun 16, 2011 #1
    If we have an exponential fuction,

    (for example)

    Limx->∞ e(x2+2x+1)/(x2-3)

    Would we first determine the limit of the "argument" (not sure if right word) of ex and then replace the "argument" with the limit and then evaluate it?

    So for the example above,

    The limit of (x2+2x+1)/(x2-3) would be 1 as it approaches ∞.

    Then we would evaluate e1.


    Would this process be correct?:smile:
     
  2. jcsd
  3. Jun 16, 2011 #2

    micromass

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    Hi BloodyFrozen! :smile:

    Yes, this is correct. In fact, for every continuous function f, we have

    [tex]\lim_{x\rightarrow a}{f(g(x))}=f\left(\lim_{x\rightarrow a}{g(x)}\right)[/tex]

    In your example, we have [itex]f(x)=e^x[/itex] and [itex]g(x)=\frac{x^2+2x+1}{x^2-3}[/itex].
     
  4. Jun 16, 2011 #3
    Ok,

    And also for limx->-∞ex2/(x-3)

    The g(x) part of f(g(x))

    equals -∞ at -∞

    so,

    e-∞ would "theorectically" approach 0 right?
     
  5. Jun 16, 2011 #4

    micromass

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    Yes, that limit is indeed 0! :smile:
     
  6. Jun 16, 2011 #5
    OK, and one last question:tongue2:

    limx->-∞ (x2+2x+1)/(-x2+4x+4)

    Could we evaluate each one seprately?

    top -> -∞
    bottom ->∞

    so "technically" could we say -∞/∞ = -1

    So would the limit be -1?

    [P.S. I do know about horizontal asymptotes (Algebra way and Calculus way)]
     
    Last edited: Jun 16, 2011
  7. Jun 16, 2011 #6

    micromass

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    Yes, the limit is -1. It seems you understand these things quite well! :smile:

    However, I must say that I don't like

    [tex]\frac{\infty}{-\infty}=-1[/tex]

    although it's clear what you mean. I wouldn't write this on a test or something. A better way of writing this would be

    [tex]\lim_{x\rightarrow -\infty}{\frac{x^2+2x+1}{-x^2+4x+1}}=\lim_{x\rightarrow -\infty}{\frac{x^2}{-x^2}}=\lim_{x\rightarrow -\infty}{-1}=-1[/tex]
     
  8. Jun 16, 2011 #7

    gb7nash

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    Yes, but not for the reason that you stated. If you're still not convinced, look at:

    limx->-∞ (x2)/(-x)

    When the highest power on the top and bottom is the same, you simply take the coefficients of the highest power and divide. In your example, it would be 1/-1 = -1
     
  9. Jun 16, 2011 #8
    Not generally. Consider lim (x)/(x^2) as x -> ∞. "Top over bottom" is ∞/∞ but it does not equal 1; it is 0.

    To solve lim (x^2+2x+1)/(-x^2+4x+4) as x -> ∞, I'd multiply the argument by [tex]\frac{1/x^2}{1/x^2}[/tex] to get
    [tex]\lim_{x \to \infty} \frac{1 + 2/x + 1/x^2}{-1 + 4/x + 4/x^2}[/tex] which is much more clear.
     
  10. Jun 16, 2011 #9
    So what we do would be factor the largest x-exponent from the denominator. That way we would get the other "parts" to m/xn (where m and n are any numbers) as it approaches +/-∞ it'd equal zero. That way it'd be x2/-x2 like you guys said. Am I correct?
     
  11. Jun 16, 2011 #10
    Yes, I know.:smile:

    I mentioned the lower leveled class (algebra?) way.

    If n/m,

    If power are same, take coefficients of largest power from numerator and denominator and divide n/m

    If n>m, there is no horizontal asymptote.

    If n<m, horizontal asymptote at y=0

    :rofl:
     
  12. Jun 16, 2011 #11

    micromass

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    Yes, that's good!
     
  13. Jun 16, 2011 #12

    micromass

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    The good thing about calculus is that you can now forget all this nonsense :biggrin: Just evaluate things like Unit did. Mathematicians don't like memorizing.
     
  14. Jun 16, 2011 #13
    Ok, Spivak's 4th Edition is doing well so far.:tongue2:

    BTW are there any "challenging" limits that I might want to be aware of?:surprised
     
  15. Jun 16, 2011 #14

    micromass

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    Oh, you want challenges, hey. :devil:

    [tex]\lim_{x\rightarrow +\infty}{\frac{x+\sqrt{x^2+3}}{-2x-\sqrt{2x^2-4}}}[/tex]

    [tex]\lim_{x\rightarrow 0}{\frac{e^x-1}{x}}[/tex]

    [tex]\lim_{x\rightarrow +\infty}{\frac{\sin(x)}{x}}[/tex]

    [tex]\lim_{x\rightarrow 0}{\frac{\cos(x)-1}{\sin(x)}}[/tex]

    There are not difficult, but they require some more thought than just plug-and-chug nonsense.
     
  16. Jun 16, 2011 #15

    Mark44

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    Good point!
     
  17. Jun 16, 2011 #16
    Yay, no -b/(2a).
    micromass, I'll attempt them tomorrow.
    No TeX on phones.

    More cont on next page:)
     
  18. Jun 16, 2011 #17
    Also, what is the first and second value in an ordered pair called? I remember they had a specific math "name"
     
  19. Jun 16, 2011 #18

    micromass

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    The abscissa and the ordinate. Although I've never used those terms in my life.
     
  20. Jun 16, 2011 #19
    Thanks to all.
    Goodnight.







    Atleast to this thread.;)
     
  21. Jun 17, 2011 #20
    I'm having trouble with 1 and 3, but I think I got 2 and 4.

    For 2 and 4, I used L'Hopital's Rule

    2. (e0-1)/0 = 0/0 ->Differentiate top and bottom
    ex = e0 = 1:biggrin:
    4. (cox(x)-1)/(sin(x)) = 0/0 Differentiate
    -sin(x)/cos(x) = -0/1 = 0:rofl:

    I can't think of any other ways to find the limits of the above and I'm stuck on numbers 1 and 3:cry:
     
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